196 PRIME Opt. Maths Book - IX Exercise 5.4 1. i) Write down the basic formula for reciprocal relations on trigonometric ratios. ii) Write down the basic formula for Pythagoras relation on trigonometric ratios. iii) Write down the basic formula for reciprocal rational on trigonometric ratios. iv) Express Cosecq in terms of Tanq. v) Express CotA in terms of SecA. 2. Prove that the followings. i) Sinq.Tanq.Secq = Tan2 q ii) Sinq × 1 Cot 2 + θ = 1 iii) Sec2 q(1 – Sin2 q) = 1 iv) Tan2 A – Sin2 A = Tan2 A.Sin2 A v) Cot2 A – Cos2 A = Cot2 A.Cos2 A 3. i) Sinq(1 + Sinq) – Sinq(1 – Cosecq + Sin2 q) = 1 ii) (1 – Sin2 q)(1 + Cot2 q) = Cot2 q iii) (1 + Tan2 A) (1 – Cos2 A) = Tan2 A iv) Sin4 q – Cos4 q = 2Sin2 q – 1 v) (Sinq + Cosq)2 = 1 + 2Sinq.Cosq 4. i) Sec4 q – Tan4 q = Sec2 q + Tan2 q ii) Cosec4 q – Cot4 q = 2Cosec2 q – 1 iii) Cos 1 – Sin 4 4 q q = 1 + 2Tan2 q iv) – Sin 1 Cos 4 4 q q = 1 + 2Cot2 q v) Sin4 q + Cos4 q = 1 + 2Sin2 qCos2 q 5. i) 1 CosecA – CotA = CosecA + CotA ii) 1 SecA + TanA = SecA – TanA iii) CosA 1 + SinA = SinA CosA 1 – iv) CosA SinA 1 – = SinA 1 + CosA v) Sec A – 1 2 ^ h 1 – Sin A2 ^ h = SinA 6. i) TanA TanA 1 1 – + = CosA SinA CosA – SinA + ii) Cot Cot 1 1 a – a + = Cos S – in Cos Sin a a a a + iii) – SinA SinA 1 1 + = SecA – TanA iv) – CosA CosA 1 1 + = CosecA + CotA v) CosA CosA 1 1 – + = (CotA – CosecA)2
PRIME Opt. Maths Book - IX 197 7. i) 1 + SinA 1 – SinA = (SecA + TanA)2 ii) 1 1 + CosA + 1 1 – CosA = 2Cosec2 A iii) 1 1 – SinA – 1 1 + SinA = 2TanA.SecA iv) 1 + SinA CosA + CosA 1 + SinA = 2SecA v) 1 – CosA SinA + SinA 1 – CosA = 2CosecA PRIME more creative questions: 8. Prove that the following identities. a. i) Sin³q – Cos³q Sinq – Cosq = 1 + SinqCosq ii) (1 + Sina + Cosa)2 = 2(1 + Sina)(1 + Cosa) iii) (1 – SinA – CosA)2 = 2(1 – SinA)(1 – CosA) iv) TanA 1 – CotA + CotA 1 – TanA = 1 + SecA.CosecA v) Sin4 A – Cos4 A SinA – CosA = SinA + CosA b. i) (1 + Cotq – Cosecq) (1 + Tanq + Secq) = 2 ii) Sec²q + Cosec²q = Tanq + Cotq iii) (SinA + CosecA)2 + (CosA + SecA)2 = Tan2 A + Cot2 A + 7 iv) Tanq Secq – 1 + Tanq Secq + 1 = 2Cosecq v) Sin6 A + Cos6 A = 1 – 3 Sin2 A.Cos2 A Answer Show to your teacher.
198 PRIME Opt. Maths Book - IX Worked out Examples 5.5 Complex identities: We discuss about the simple trigonometrical identities in previous exercise where as we introducing some difficult and technical trigonometrical identities in this chapter. Hence, We try to explain the procedures to prove them through some worked out examples. 1. Prove that: CotA TanA 1 – + TanA CotA 1 – = 1 + SecA. CosecA. L.H.S. = cot A TanA 1 – + TanA CotA 1 – = SinA CosA CosA SinA 1 – + CosA SinA SinA CosA 1 = SinA SinA CosA CosA SinA – + CosA CosA SinA SinA CosA – = CosA SinA × SinA CosA SinA – – SinA CosA × SinA CosA CosA – = Sin²A – Cos²A CosA.SinA(SinA – CosA) = (SinA – CosA)(Sin²A + SinA.CosA + Cos²A CosA.SinA(SinA – CosA) = CosA SinA 1 + SinA CosA = CosA SinA 1 + SinA CosA SinA CosA = SecA. CosecA + 1. = R.H.S. is proved 2. Sin2 A Cos2 B – Cos2 A Sin2 B = Sin2 A – Sin2 B L.H.S. = Sin2 A Cos2 B – Cos2 A Sin2 B = Sin2 A (1 – Sin2 B) – (1 – Sin2 A) Sin2 B = Sin2 A – Sin A Sin B 2 2 – Sin2 B + Sin A Sin B 2 2 = Sin2 A – Sin2 B = R.H.S. Proved.
PRIME Opt. Maths Book - IX 199 3. SecA – TanA – 1 SecA – TanA + 1 = 1 – SecA – TanA 1 + SecA + TanA L.H.S. = SecA – TanA – 1 SecA – TanA + 1 = (SecA – TanA)–(Sec²A – Tan²A) (SecA – TanA)+(Sec²A – Tan²A) = (SecA – TanA) – {(SecA + TanA)(SecA – TanA)} (SecA – TanA) + {(SecA + TanA)(SecA – TanA)} = (SecA – TanA)(1 – SecA – TanA) (SecA – TanA)(1 + SecA + TanA) = 1 – SecA – TanA 1 + SecA + TanA = R.H.S. is proved 4. CosecA + CotA – 1 1 – CosecA + CotA = SinA 1+ CosA L.H.S = CosecA + CotA – 1 1 – CosecA + CotA = (CosecA + CotA) – (Cosec²A + Cot²A) 1 – CosecA + CotA = (CosecA + CotA) – {(CosecA + CotA) (CosecA – CotA)} 1 – CosecA + CotA = (CosecA + CotA)(1 – CosecA + CotA) 1 – CosecA + CotA = SinA 1 + SinA CosA = SinA 1 + CosA = R.H.S. Proved 5. (3 – 2Sin2 q) (2Cot2 q – 3) = (1 + 3Cot2 q) (2 – 5Sin2 q). L.H.S. = (3 – 2Sin2 q) (2Cot2 q – 3) = (3 – 2sin2 q) Sin 2Cos 2 – 3 2 i i c m = (3 – 2Sin2 q)( 2Cos²q – 3Sin²q Sin²q ) = ( 3 Sin²q – 2Sin²q Sin²q ){2(1 – Sin2 q) – 3Sin2 q} = {3 (1 + Cot2 q) – 2} {2 – 2Sin2 q – 3Sin2 q} = (3 + 3Cot2 q – 2) (2 – 5Sin2 q) = (1 + 3Cot2 q) (2 – 5 Sin2 q) = R.H.S. Proved.
200 PRIME Opt. Maths Book - IX Exercise 5.5 1. i) (Cosq + Sinq) (Secq + Cosecq) = 2 + Secq.Cosecq ii) (1 + Sinq – Cosq)2 + (1 – Sinq + Cosq)2 = 4(1 – Sinq.Cosq) iii) 1 – Cot2 a + Cot4 a = Sin2 a(1 + Cot6 a) iv) (CosecA – SinA)(SecA – CosA)(TanA + CotA) = 1 v) Cot2 A Sin2 B – Cosec2 A Tan2 B = Cot2 A – Cot2 B 2. Prove that the followings: i. Sin2 aCos2 b – Cos2 aSin2 b = Cos2 b – Cos2 a. ii. Cos B Tan A 2 2 – Cos A Tan B 2 2 = Tan2 A – Tan2 B. iii. Cot2 qCosec2 b – Cosec2 q . Cot2 b = Cosec2 q – Cosec2 b iv. (xCosa + ySina)2 + (xsina – yCosa)2 = x2 + y2 . v. (1 + 2Cosq)2 = 1 + 3Cosq – 4Cos2 q 1 – Cosq 3. Prove that the followings: i. TanA CotA 1 – + CotA TanA 1 – = SecA.CosecA + 1 ii. Tan – Tan 1 2 a a + Cot2 a Cota – 1 = Seca.Coseca + 1. iii. Cot Sin 1 – 2 a a + Tan Cos 1 – 2 a a = Sina.Cosa + 1. iv. (3 – Tan2 q) (2 – 3Sin2 q) = (3 – 4Sin2 q) (2 – Tan2 q) v. (3 – 2Cot2 q) (2 – Cos2 q) = (3 – 5Cos2 q) (2 + Cot2 q) 4. Prove that the followings: i. 1 + Secq – Tanq 1 + Secq + Tanq = – Cos 1 Sin q q ii. Secq + Tanq – 1 1 – Secq + Tanq = Cos 1 Sin i + i iii. 1 – Cosecq + Cotq Cosecq + Cotq – 1 = Sin 1 – Sin i i iv. Cosecq + Cotq – 1 1 – Cosecq + Cotq = Cos Sin 1 – i i v. 1 – Secq + Tanq Secq – Tanq – 1 = Sin Cos 1 – i i 5. Prove that the followings: i. Cosecq + Cotq + 1 Cosecq – Cotq – 1 = 1 + Cosecq + Cotq 1 – Cosecq – Cotq ii. 1 – Secq – Tanq 1 + Secq + Tanq = Secq – Tanq – 1 Secq + Tanq + 1
PRIME Opt. Maths Book - IX 201 iii. 1 CosecA – CotA – SinA 1 = SinA 1 – 1 CosecA + CotA iv. Cos 1 i – Sec Tan 1 i i + = Sec Tan 1 i i – – Cos 1 i v. SinA CosA 1 – + SinA CosA 1 + = 2TanA 6. PRIME more creative questions: a. i. Sin6 A – cos6 A = (1 – 2Cos2 A)(1 – Cos2 A + Cos4 A) ii. Sin8 q – Cos8 q = (2Sin2 q – 1)(1 – 2Sin2 q + 2Sin4 q) iii. (1 + Cotq – Cosecq) (1 + Tanq + Secq) = 2 iv. Sec Tan a – 1 a – Cos Sin 1 a a + = 2Cota v. SinA SinA 1 1 – + – SinA SinA 1 1 – + = TanA vi. TanA Tan A – 1 2 + TanA CotA 1 – = 1 + SecA.CosecA b. i. Sec Tan Sec Tan 2 2 6 6 i i i i + + = 1+Sec2 q.Tan2 q ii. 2(Sin6 A+Coc6 A) – 3(Sin4 A+Cos4 A) + 1 = 0 iii. Tan3 a+Cot3 a = Sec3 aCosec3 a-3Seca.Coseca iv. SinA CosA SinA CosA 1 1 – – + + = SecA + TanA v. CosA SinA CosA SinA 1 1 – – + + = SinA 1 + CosA c. i. (3 – 4Sin2 a) (1 – 3Tan2 a) = (3 – Tan2 a)(1 – 4Sin2 a) ii. Sinx Cosy Cosx + + Siny Cosx Cosy – = Sinx Cosy Cosx – + Siny Cosx Cosy + iii. 1 + Sec2 A.Cot2 C 1 + Sec2 B.Cot2 C = 1 + Tan2 A.Cos2 C 1 + Tan2 B.Cos2 C iv. 1 + SinA – CosA 1 + SinA + CosA = 1 + SinA + CosA 1 + SinA – CosA = 2 CosecA v. Sin Sin Cos Cos i b – i b + + Cos Cos Sin Sin i b – i b + = 0 Answer Show to your subject teacher.
202 PRIME Opt. Maths Book - IX 5.6 Use of given trigonometrical ratio to find the rest ratios: In this chapter, we discuss the process of finding all the trigonometrical ratios, if one of them is given either directly or in the form of algebric expression. 1. Find the ratio of Sinq and Cotq from the given right angled triangle. Solution : In right angled triangle ABC \B = 90° BC = 4cm = b AC = 5cm = h AB = p = h b –2 2 = 5 4 –2 2 = 9 = 3cm Then, Sinq = h p = AC BC = cm cm 5 4 = 5 4 Cotq = p b = BC AB = cm cm 4 3 = 4 3 2. If Sin A= 5 3 , find the vale of Sec2 A – Tan2 A. Solution : In right angled DABC \B = 90° \A = reference angle SinA = 5 3 or h p = 5 3 Here, p = 3cm h = 5 cm Then, b = h p –2 2 = 5 3–2 2 = 16 = 4 cm Again, Sec2 A – Tan2 A = – b h b 2 p 2 a k ` j = – 4 5 4 3 2 2 a k ` j = 16 25 – 9 = 16 16 = 1 Worked out Examples q B A C 4cm 5cm
PRIME Opt. Maths Book - IX 203 3. If a Tanθ = b �ind the value of Sin Cos Sin – Cos 2 2 2 2 i i i i + Solution: or, aTanθ = b or, Tanθ = a b or, b p = a b p = b b = a h = p b 2 2 + = a b 2 2 + Then, sin cos sin – cos 2 2 2 2 i i i i + = h p h b h p h b – 2 2 2 2 ` + ` a j a j k k = ² ² – a b b a b a a b b a b a ² ² ² ² ² ² 2 2 2 2 + + + + + c c a m a m k k = a b b – a 2 2 2 2 + × b a a b 2 2 2 2 + + = b a b – a 2 2 2 2 + 4. Express all the trigonometrical ratios in terms of SinA. Solution: Let, Sin A = K h p = k 1 p = k h = 1 b = h – p 2 2 = 1 – k2 Then, Sin A = Sin A Cos A = h b = k 1 1 2 - = 1 – Sin A 2 Tan A = b p = k k 1 – 2 = Sin A SinA 1 – 2 Cosec A = p h = k 1 = SinA 1 Sec A = b h = 1 k 1 – 2 = 1 sin A 1 – 2 Cot A = p b = k 1 – k2 = sin sin A 1 – A 2
204 PRIME Opt. Maths Book - IX 5. If SinA – CosA = 0, find the value of SecA. Solution : SinA – CosA = 0 or, SinA = CosA or, CosA SinA = 1 or, TanA = 1 or, b p = 1 \ p = 1 b = 1 h = p b 2 2 + = 2 Then, SecA = b h = 1 2 = 2 SecA = 2 6. If 2cos2q – Sinq – 1 = 0, find the value of sinq. Solution : 2cos2 q – Sinq – 1 = 0 or, 2(1 – Sin2 q) – Sinq – 1 = 0 or, 2 – 2Sin2 q – Sinq – 1 = 0 or, 2Sin2 q + Sinq – 1 = 0 or, 2Sin2 q + (2 – 1)Sinq – 1 = 0 or, 2Sin2 q + 2sinq – Sinq – 1 = 0 or, 2Sinq(Sinq + 1) – 1(Sinq + 1) = 0 or, (Sinq + 1) (2Sinq – 1) = 0 Either OR Sinq + 1 = 0 2Sinq – 1 = 0 \ Sinq = – 1 Sinq = 2 1 \ Sinq = – 1 or 2 1 7. If 5cosq + 12 Sinq = 13, find the value of Tanq. Solution : 5Cosq + 12Sinq = 13 or, 5Cosq = 13 – 12 Sinq Squaring on both sides, or, 25Cos2 q = 132 – 312 Sinq + 144Sin2 q or, 25 – 25Sin2 q = 169 – 312 Sinq + 144Sin2 q
PRIME Opt. Maths Book - IX 205 or, 169Sin2 q – 312Sinq + 144 = 0 or, (13Sinq – 12)2 = 0 or, 13Sinq – 12 = 0 or, Sinq = 13 12 \ Sinq = 13 12 or, h p = 13 12 \ p = 12 h = 13 \ b = h p –2 2 = 169 1– 44 = 5 \ Tanq = b p 5 12 = Alternative method 5cosq + 12sinq = 13 Dividing both sides by cosq or, Cos 5Cos i i + Cos 12Sin i i = Cos 13 i or, 5 + 12Tanq = 13Secq Squaring on both sides, or, (5 + 12Tanq)2 = (13Secq)2 or, 25 + 120Tanq + 144Tan2 q = 169 + 169Tan2 q or, 25Tan2 q – 120Tanq + 144 = 0 or, (5Tanq – 12)2 = 0 or, 5Tanq – 12 = 0 \ Tanq = 5 12 8. If Sinq + Cosq = 2 Sinq, prove that Sinq – Cosq = ± 2 Cosq. Solution : Sinq + Cosq = 2 Sinq Squaring on both sides, or, (Sinq + Cosq)2 = ( 2 Sinq)2 or, Sin2 q + 2SinqCosq + Cos2 q = 2Sin2 q or, 1 + 2SinqCosq = 2Sin2 q or, 2SinqCosq = 2Sin2 q – 1 or, – 2SinqCosq = 1 – 2Sin2 q or, 1 – 2SinqCosq = 2 – 2Sin2 q or, Sin2 q + Cos2 q – 2SinqCosq = 2(1 – Sin2 q) or, (Sinq – Cosq)2 = 2Cos2 q \ Sinq – Cosq = ± 2 Cosq is proved
206 PRIME Opt. Maths Book - IX Exercise 5.6 1. i) Is a triangle having sides 6cm, 8cm and 10cm a right angled? ii) If Sinq – 1 = 0, find Cosq. iii) If Sinq – Cosq = 0, find Sinq. iv) If Tana = k, what will be the value of Cosa? v) If SecA = CosecA, find the value of CotA. 2. Find the trigonometric ratios of Tana, Cosa Cosecq and Cotq from the following diagrams. i) K L M 10cm 6cm a q ii) 8cm 17cm P Q R q a iii) S 17cm 7cm 7 2cm Q P a q iv) 13cm 4cm 3cm B C a q D A v) C B A D q a 13cm 12cm 3cm 3. i) If Sin A = 5 3 , find the value of TanA and Sec A. ii) If Cos A = 13 5 , find the value of Cosec2 A – Cot2 A. iii) If 4Tan A = 3, find the value of 5(Sin A + Cos A) iv) If 17 Sin A = 15, find Sec2 A – Tan2 A. v) If 3 TanA = 4, frind the value of CosA SinA CosA SinA 2 3 – +
PRIME Opt. Maths Book - IX 207 4. i) If n Tan A = m, find the value of Sin2 A + Cos2 A. ii) If p CosA = q, prove that p q –2 2 Cosec A = p. iii) If 1 – Cos q = 2 1 , find the value of Cotq – Tanq. iv) If Sinq – Cosq = 0, find Cosecq. v) If Secq – Cosecq = 0, find the value of Secq. 5. i) Express all the trigonometrical ratios in terms of Cos A. ii) Express all the trigonometrical ratios in terms of Tan A. iii) Express all the trigonometrical ratio in terms of Sin A. iv) Express all the trigonometrical ration in terms of Cosec A. v) If 3CosA = 4, find all other trigonometrical ratios. 6. i) If Sinq = a a 1 – 1 + , find cot2 q – Cosec2 q. ii) If 41CosA = 40, find Tan A TanA 1 – 2 iii) If CosA = a b a 2 2 + , find CotA + TanA iv) If (m2 + n2 )SinA = m2 – n2 , prove that (m2 – n2 )CotA = 2mn. v) If yTanq = x, find the value of xSin yCos xSin – yCos i i i i + 7. i) If 5TanA = 4, prove that SinA CosA SinA CosA 2 5 3– + = 14 5 ii) If a Cotq = b, prove that aSinq + bCosq = a b 2 2 + iii) If SinA = 5 3 and SinB = 13 12 , find the value of SinACosB + CosASinB. iv) If ( 3 1 + ) Tanq = 3 1– , Find the value of Sinq + Cosq. v) If x y 2 2 + Cosq = x, find the value of xSinq – yCosq. 8. PRIME more creative questions: a. i) If 2Sin2 q = 2 + Cosq, find the value of Cosq. ii) If 2Sin2 q – Cosq – 1 = 0, find the value of Cosq where ‘q’ lies in first quadrant. iii) If Secq.Tanq = 2 , find the value of Sinq. iv) If 2 3 Cos2 q = Sinq find the value of Sinq where q lies in first quardrent. v) If Tan2 q + ( 3 – 3 1 ) Tanq = 1, find the value of Tanq where ‘q’ lies in first quadrent. b. i) If 3Cosq + 4Sinq = 5, find the value of Sinq. ii) If 4Cosq + 3Sinq = 5, find the value of Cotq. iii) If 8Sinq + 15Cosq = 17, find the value of Tanq. iv) If Cota + Coseca = m, find the value of Cosa where ‘a’ lies in first quadrent. v) If Cos4 A + Cos2 A = 1, prove that Tan4 A + Tan2 A = 1
208 PRIME Opt. Maths Book - IX Answer 1. Show to your teacher. 2. i) Tana = 3 4 , Cosa = 5 3 , Cosecq = 3 5 , Cotq = 3 4 ii) Tana = 15 8 , Cosa = 17 15 , Cosecq = 17 15 , Cotq = 15 8 iii) Tana = 24 7 , Cosa = 25 24 , Cosecq = 2 1 , Cotq = 1 iv) Tana = 3 4 , Cosa = 5 3 , Cosecq = 13 5 , Cotq = 5 12 v) Tana = 12 5 , Cosa = 13 12 , Cosecq = 5 3 , Cotq = 3 4 3. i) TanA = 4 3 , SecA = 4 5 ii) 1 iii) 7 iv) 1 v) 11 5 4. i) 1 iii) 3 2 iv) 2 v) 2 5. i) Show to your subject teachers. 6. i) –1 ii) 1519 360 iii) ab a b 2 2 + v. x y x y – 2 2 2 2 + 7. iii) 65 63 iv) 2 3 v) o 8.a. i) 0, 2 1 ii) 2 1 iii) 2 1 iv) 2 3 v) 3 1 b. i) 5 4 ii) 3 4 iii) 15 8 c. iv) 9 v) 2 1 c. i) If CosA + SinA = 2 CosA, prove that CosA – SinA = ± 2 SinA. ii) If Cotq + Cosq = a and Cotq – Cosq = b, prove that a2 – b2 = 4 ab iii) If Secq + Tanq = a, prove that (a2 + 1) sinq = a2 – 1. Where ‘q’ lies in first quadent. iv) If Tanq + Cotq = 3, find the value of Sec2 q + Cosec2 q. v) If TanA = 1 in a right angled DABC, find the value of SinACosA.
PRIME Opt. Maths Book - IX 209 5.7 Trigonometrical Ratios of some standard angles The trigonometrical ratios of some standard angles 0°, 30°, 45°, 60° & 90° are discussed and geometrical interpretations are given in this chapter. i) Trigonometrical ratios of 0°. q B A C In the right angled DABC, \B = 90° \C = q (reference angle) If A approaches to B, the value of q decreases and at q = 0 ‘A’ coincides with ‘B’ such that AB = 0 and AC = BC Sinq = h p = AC AB ( Sin0° = BC 0 = 0 Cosq = h b = AC BC ( Cos0° = BC BC = 1 Tanq = b p = BC AB ( Tan0° = BC 0 = 0 Also, Cosec0° = ∞, Sec0° = 1, Cot0° = ∞ ii) Trigonometrical ratios of 30° & 60°. Let, ABC is an equilateral triangle In an equilateral DABC, AD⊥BC AB = BC = AC = 2a (say) BD = DC = a \A = \B = \C = 60° AD = AB – BD 2 2 = 2 – a a 2 2 ^ ^ h h = a 3 Then, In rt. \ed DADB, \BAD = 90° – 60° = 30° AB = 2a = h BD = a = p AD = a 3 = b B A D 60° 30° 2a 2a a a C
210 PRIME Opt. Maths Book - IX Sin30° = h p = AB BD = a a 2 = 2 1 Cos30° = h b = AB AD = a a 2 3 = 2 3 Tan30° = b p = AD BD = a a 3 = 3 1 Also, Cosec30° = 2 Sec30° = 3 2 Cot30° = 3 Again, In rt. \ed DABD, \ABD = 60° AD = a 3 = p BD = a = b AB = 2a = h Sin60° = h p = AB AD = a a 2 3 = 2 3 Cos60° = h b = AB BD = a a 2 = 2 1 Tan60° = b p = BD AD = a a 3 = 3 Also, Cosec60° = 3 2 Sec60° = 2 Cot60° = 3 1 iii) Trigonometrical ratios of 45° B A C Let DABC is an isoceles right angled triangle where \B = 90° AB = BC = a (say) \ \A = \C = 45° Then \ AC = AB BC 2 2 + = a a 2 2 + = a 2
PRIME Opt. Maths Book - IX 211 Also, Taking reference angle A. Sin45° = h p = AC BC = a a 2 = 2 1 Cos45° = h b = AC AB = a a 2 = 2 1 Tan45° = b p = AB BC = a a = 1 Cosec45° = p h = BC AC = a a 2 = 2 Sec45° = b h = AB AC = a a 2 = 2 Cot45° = p b = BC AB = a a = 1 iv) Trigonometrical ratios of 90°. Let, ABC is a right angled triangle. Where, \B = 90° \C = Reference angle = q When C approaches to B, then reference angle increases and at q = 90°, C coincides with B. Such that BC = 0 AB = AC Now, SinC = h p = AC AB & Sin90° = AB AB = 1 CosC = h b = AC BC & Cos90° = AC 0 = 0 TanC = b p = BC AB & Tan90° = AB 0 = ∞ Also, Cosec90° = 1 Sec90° = ∞ Cot90° = 0 Table for the values w.r.t. angles. B A C Write down = 0, 1, 2, 3, 4 Dividing by 4 = 4 0 , 4 1 , 4 2 , 4 3 , 4 4 Square root = 4 0 , 4 1 , 4 2 , 4 3 , 4 4 Result = 0, 2 1 , 2 1 , 2 3 , 1
212 PRIME Opt. Maths Book - IX Tabulation of the above values respectively. Angles Ratios 0° 30° 45° 60° 90° Sin 0 2 1 2 1 2 3 1 Cos 1 2 3 2 1 2 1 0 Tan 0 3 1 1 3 ∞ Cosec ∞ 2 2 3 2 1 Sec 1 3 2 2 2 ∞ Cot ∞ 3 1 3 1 0 To remember Sin0° = Cos90° = Tan0° = Cot90° = 0 Sin90° = Cos0° = Tan45° = Cosec90° = Sec0° = Cot45° = 1 Sin30° = Cos60° = 2 1 Sin60° = Cos30° = 2 3 Tan30° = Cot60° = 3 1 Tan60° = Cot30° = 3 Cosec30° = Sec60° = 2 Cosec60° = Sec30° = 3 2
PRIME Opt. Maths Book - IX 213 1. Find the value of sin30°.Cot60°.Cos30°.cosec30° Solution: Here, Sin30°.Cot60°.Cos30°.Cosec30° = 2 1 3 1 × × × 2 3 2 = 2 1 2. 3Tan2 30° + Sin2 60° + Cos2 45° Solution: Here, 3Tan2 30° + Sin2 60° + Cos2 45° = 3 × 3 1 2 c m + 2 3 2 1 2 2 c + c m m = 3 × 3 1 4 3 2 1 + + = 4 4 3 + + 2 = 4 9 = 2 4 1 3. Sin2 p 6 + 2 1 Sec2 p 3 + 2Tan2 p 4 + Cosec2 p 2 Solution : Here, Sin2 p 6 + 1 2 Sec2 p 3 + 2Tan2 p 4 + Cosec2 p 2 = Sin2180° 6 + 1 2 Sec2180° 3 + 2Tan2180° 4 + Cosec2180° 2 = Sin2 30° + 1 2 Sec2 60° + 2Tan2 45° + Cosec2 90° = 2 1 2 a k + 1 2 (2)² + 2(1)² + (1)² = 1 4 + 2 + 2 + 1 = 4 1 2 + 0 = 4 21 = 5 1 4 Worked out Examples
214 PRIME Opt. Maths Book - IX Exercise 5.7 1. i) If Sinq = 2 3 , find Cosq. ii) In an isosceles DABC, ∠B = 90°. What is the value of Tan 45°. iii) Find the value of Tan2 60. iv) Find the value of Sin2 60° + Cos2 60°. v) Find the value of Sec2 30° – Tan2 30° 2. Find the value of the followings. i) Sin0° + Cos90° + Tan45° ii) Sin90°.Cos60°.Cosec30° iii) Tan45° + 2Sin60° – 2Cos60° iv) Sec0°.Cosec45°.Cos45° v) 2 3 Tan60°.Cot30°.Cos30° 3. Find the value of the followings. i) Sin2 30° + Cos2 45° + Tan2 60° ii) 4Cos2 30° + 3Cot2 60° + Cosec2 45° iii) Cos2 0° + Sin2 90° + Sec2 45° iv) Sin p 3 .Cos p 6 .Tan p 4 v) 3 1 Cos2 p 6 + 4Sin2 p 3 + Cot2 p 4 4. Prove that the following. i) Cos2 p 2 + Sin2 p 4 + 2 1 Tan2 p 4 = Sin2 p 2 ii) Tan Tan 1 30 1 30 ° – ° + = 2 – 3 iii) Sin Cos 1 3 1 6 – r r + = 7 + 4 3 4. If A = 30°, B = 60°, C = 45° & D = 90°, find the value of SinA + Cos2 B + Tan2 C.CosecD. Solution: Here, SinA + Cos2 B + Tan2 C.CosecD = Sin 30° + Cos2 60° + Tan2 45°.Cosec90° = 2 1 + 2 1 2 a k + 1 2 ^ h × 1 = 4 2 1 + + 4 = 1 4 3 5. If 2SinA = 1, find the value of A. Solution: Here, 2SinA = 1 or, SinA = 2 1 or, SinA = Sin30° \ A = 30°
PRIME Opt. Maths Book - IX 215 Answer 1. Show to your teacher. 2. i) 1 ii) 1 iii) 3 iv) 1 v) 3 3. i) 3 4 3 ii) 6 iii) 4 iv) 4 3 v) 4 4 1 5. i) 2 3 + 1 ii) 0 iii) 2 4 3 iv) 2 2 1 v) 3 6. i) 60° ii) 60° iii) 30° iv) 0° v) 30° 7. i) 30° ii) 4 iv) Cot60°.Cot30° + 1 Cot30° – Cot60° = Tan60° v) Sin30°Cos60° + Cos30°Sin60° = 1 5. If A = 0°, B = 30°, C = 45°, D = 60°, �ind the value of followings. i) SinA + 2CosB + CotC + TanD ii) 3 TanB + CotC – 2 3 CotD iii) Sin2 B – Cos2 C + Tan2 D iv) 3Cot2 D – 3 2 Cos2 B + 4Sin2 C v) Cos(A + B) + Cos(D – B) 6. Find the value of A from the followings. i) 2CosA – 1 = 0 ii) 2SinA = 3 iii) 3 TanA – 1 = 0 iv) CosA – 1 = 0 v) CotA – 3 = 0 7. PRIME more creative questions. i) If 3Tanθ – 3 = 0, �ind the value of ‘θ’. ii) Evaluate : 13 + 7 + 1 + 2 3Sin60° iii) Prove that : Sin Sin 1 30 1 60 – ° – ° = Tan Tan 1 30 1 30 ° – ° + iv) Geometrically prove that Sin60° = 2 3 v) Geometrically prove that Cos45° = 2 1
216 PRIME Opt. Maths Book - IX 5.8 Trigonometric ratio of complementary angles and solution of right angled triangle Let us consider that DPQR is a right angled triangle where, \Q = 90° \R = q (refrence angle) \ \P = 90° – q (Complementary angle of q) Then, for the refrence angle (90° – q) Sin(90° – q) = h p PR QR = = Cosq Cos(90° – q) = h b PR PQ = = Sinq Tan (90° – q) = b p PQ QR = = Cotq Cosec(90° – q) = p h QR PR = = Secq Sec(90° – q) = b h PQ PR = = Cosecq Cot (90° – q) = p b QR PQ = = Tanq 1. Find the value of : Sin25° – Cos65° Solution: Here, Sin25° – Cos65° = Sin25° – Cos(90° – 25°) = Sin25° – Sin25° = 0 2. Prove that : Cos55° + Sin65° = Sin35° + Cos25° Solution : L.H.S. = Cos55° + Sin65° = Cos(90° – 35°) + Sin(90° – 25°) = Sin35° + Cos25° = R.H.S. proved 3. Prove that : Sin(90° – q) Cosq . Cosq Sin(90° – q) . 1 Sec(90° – q) = Sinq Solution : L.H.S. = Sin(90° – q) Cosq . Cosq Sin(90° – q) . 1 Sec(90° – q) = Cosq Cosq × Cosq Cosq × 1 Cosecq Worked out Examples 90° – q q Q P R
PRIME Opt. Maths Book - IX 217 = 1 Cosecq = Sinq = R.H.S. proved 4. Find the side QR of right angled DPQR. Solution : In rt. \ed DPQR, \Q = 90°, \R = 60° (refrence angle) PR = 15 cm, QR = ? We have, Cos60° = h b = PR QR or, 2 1 = QR 15 or, 2QR = 15 \ QR = 7.5 cm 5. If an electric pole is tied with a rope of length 20m where the rope made 60° angle with the ground. How high does the rope from the ground? Solution: Let, AB be the height of a pole. AC be the length of the rope. Given : \B = 90° \C = 60° AC = 20 m AB = ? Now, In rt. \ed DABC, Sin60° = h p = AC AB or, 2 3 = AC AB or, 2AB = 20 3 or, AB = 2 20 3 \ AB = 17.32 m 15 cm 60° Q P R 20m C A B 60° ?
218 PRIME Opt. Maths Book - IX Exercise 5.8 1. i) What is the complementary angle of 70°. ii) Express Sin25° in terms of Cos as the complementary angle. iii) Write down the equivalent of Cos p 8 as the complementary angle. iv) Find the value of Cotq.Cot(90° – q) v) Find the value of Tan25°Tan65°. 2. Simplify : i) Sin10° – Cos80° ii) Cos70° – Sin20° – Sin40° + Cos50° iii) Cos(90° – q).Tan (90° – q).Cosec (90° – q) iv) Cot(90° – q). Sec(90° – q) Cosec(90° – q) v) Tan9°. Tan18°.Tan72°.Tan81° 3. Prove that the followings. i) Sin34° = Cos56° ii) Cot17° = Tan 73° iii) Cos40° + Sin55° = Sin50° + Cos35° iv) Sin74° – Cos37° = Cos16° – Sin53° v) Sec80° + Cosec40° = Cosec10° + Sec50° 4. Prove that the followings. i) Sin(90° – q).Cosq + Cos(90° – q).Sinq = 1 ii) Secq.Cosec(90° – q) – Cot (90° – q).Tanq = 1 iii) Tanq.Tan(90° – q) + Sinq.Sec(90° – q) = 2 iv) Sin( ) Sin 90° – i i . ( ) Cos Cos 90° – i i = Tan2 q v) ( ) ( ) Sec Cos 90 90 ° – ° – i i . ( ) ( ) Cot Tan 90 90 ° – ° – i i . Cosec(90° – q) = Cosq 5. Find the side AB from the following right angled DABC. i) B 10cm A C 30° ii) B 12cm A C 60° iii) C A B 45° 20cm iv) B A C 30° 40cm
PRIME Opt. Maths Book - IX 219 v) C A D 60° 25cm B PRIME more creative questions: 6. Find the followings. i) The angle made by ladder 20m long to the ground at the foot of the ladder is 30° which is taken against a wall. Find the height of the wall. ii) An electric pole is of height 16ft is tied with a metallic rope where the rope makes an angle of 60° with the ground, find the length of the rope. iii) A pole of height 12ft forms the shadow during the sun’s altitude of 45°, find the length of the shadow of the pole. iv) Find the height of the temple given in diagram. 2m 20 3m 60° ? v) If the kite is 102m high from the ground, find the length of the string used in the kite given in diagram. Answer 1. i) 20° ii) Cos65° iii) Sin3p 8 iv) 1 v) 1 2. i) 0 ii) 0 iii) 1 iv) 1 v) 1 4. i) 5cm ii) 6cm iii) 20cm iv) 80cm v) 50cm 5. i) 10m ii) 32ft iii) 12ft iv) 62m v) 200m 2m ? 30° 102m
220 PRIME Opt. Maths Book - IX 5.9 Trigonometrical ratios of additional angels: Other than standard angles some of the angles lie in the four quadrants like 90°-A, 90°+A, 180°-A, 180°+A, 270°-A, 270°+A, 360°-A, 360°+A and (-A). Here, we discuss the trigonometric ratios of those angles and their application in different identities. Y Y’ X –A A 180° + A 180° – A 90° + A 90° – A X’ • First quadrant : OP, PM and OM all are positive in first quadrant due to which all trigonometric ratios are positive. • Second quadrant : p = PM = +ve and b = OM = -ve, OP = +ve It makes, Sinq = h p = +ve , Cosecq = +ve Cosq = h b = -ve , Secq = -ve Tanq = b p = -ve , Cotq = -ve • Third quadrant : p = PM = -ve b = OM = -ve It makes, Sinq = h p = -ve , Cosecq = -ve Cosq = h b = -ve , Secq = -ve Tanq = b p = +ve , Cotq = +ve O M A P M O Y’ Y X’ X A q P O M Y’ Y X’ X q P
PRIME Opt. Maths Book - IX 221 • Fourth quadrant : p = PM = -ve b = OM = +ve It makes, Sinq = h p = -ve , Cosecq = -ve Cosq = h b = +ve , Secq = +ve Tanq = b p = -ve , Cotq = -ve i. Trigonomertical ratios of (90°-A) Let, XX’ and YY’ are intersected at the center of a circle ‘O’. A revolving line OP makes an angle ‘A’ with y-axis. ie. \POX = A \YOQ = A \QOX = 90° – A Draw PM⊥OX and QN⊥OX. Then, In right angled DPOM & DQON. \OQN = \YOQ = A [Alternative angles] \ \QON = 90° – A i. \PMO = \QNO = 90° ii. OM = QN iii. OP = OQ = radii of same circle. \ DPOM b DQNO (By AAS axiom) \ PM = ON and OM = QN. Then, In right angled DQON. Sin (90°-A) = OQ QN = OP OM = CosA Cos (90°-A) = OQ ON = OP PM = SinA Tan (90°-A) = ON QN = PM OM = CotA Cosec (90°-A) = QN OQ = OM OP = SecA Sec (90°-A) = ON OQ = PM OP = CosecA Cot (90°-A) = QN ON = OM PM = TanA ii. Trigonomertical ratios of (90°+A). Let, A revolving line OP makes an angle ‘A’ with OX and again ‘A’ with OY at the position of OQ. ie. \POX = A \QOY= A \QOX = 90°+A O M Y’ Y X’ X q P O N M Y’ Y X P Q X’ P 90°–A A
222 PRIME Opt. Maths Book - IX Draw the perpendiculars, PM⊥OX and QN⊥OX’. \ \OQN = \QOY = A Now, In DPMO and DQNO, i. \PMO = \QNO = 90° ii. \PAM = \OQN = A [Alternative angles] iii. OP = OQ = radii of same circle. \ DPOM ≅ DQNO (By AAS axiom) \ PM = –ON and OM = QN Then, In right \ed DQNO, Sin (90° + A) = OQ QN = OP OM = CosA Cos (90° + A) = OQ ON = OP –PM = –SinA Tan (90° + A) = ON QN = PM OM – = –CotA (and so on for others) iii. Trigonomertical ratios of (180°-A) Let, A revolving line OP makes an angle of ‘A’ with OX and again ‘A’ with OX’ at the position of OQ. ie. \POX = A \QOX’= A \ \QOX = 180°-A Draw the perpendiculars, PM⊥OX and QN⊥OX’ Now, In right angled DPMO and DQNO, i. \PMO = \QNO = 90° ii. \POM = \QON = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ PM = QN and OM = –ON. Then, In right angled DQNO, Sin (180°–A) = OQ QN = OP PM = SinA Cos (180°–A) = OQ ON = OP –OM = –CosA Tan (180°–A) = ON QN = OM PM – = –TanA (and so on for the others) N O M Y’ Y X P Q X’ A 90°+A A A N O M Y’ Y X Q P X’ 180°–A A A
PRIME Opt. Maths Book - IX 223 iv. Trigonomertical ratios of (180°+A) Let, A revolving line OP makes an angle ‘A’ with the line OX and again ‘A’ with OX’ at the position of OQ. ie. \POX = A \QOX’ = A \ \QOX = 180°+A Draw the perpendiculars, PM⊥OX and QN⊥OX’ Now, In DPMO and DQNO, i. \PMO = \QNO = 90° ii. \PM = \QON = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ OM = ON and PM = -QN. Then, In right angled DQNO, Sin (180°+A) = OQ QN = OP –PM = -SinA Cos (180°+A) = OQ ON = OP –OM = -CosA Tan (180°+A) = ON QN = OM PM – – = TanA (and so on for the others) v. Trigonomertical ratios of (270°-A) Let, A revolving line OP makes an angle of ‘A’ with OX and again ‘A’ with OY’ at the position X’ of OQ. ie. \POX = A \QOX’ = A \ \QOX = 270°-A Draw the perpendiculars, PM⊥OX and QN⊥OX’ \ \OQN = \QOY’ = A Now, In DPMO and DQNO, i. \PMO = \QNO = 90° ii. \POM = \NQO = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ PM = –ON and OM = –QN. Then, In right angled DQNO, Sin (270°-A) = OQ QN = OP –OM = –CosA Cos (270°-A) = OQ ON = OP –PM = –SinA Tan (270°-A) = ON QN = PM OM – – = CotA (And so on for the others) O M N Y’ Y X P Q X’ 180°–A A A O M Q N Y’ Y X P X’ 270°–A A A A
224 PRIME Opt. Maths Book - IX vi. Trigonomertical ratios of (270°+A) Let, A revolving line OP makes an angle ‘A’ with OX and again ‘A’ with OY’ at the position of OQ. ie. \POX = A \QOY’ = A \ \QOX(reflex) = 270°+A Draw the perpendiculars, PM⊥OX and QN⊥OX \ \OQN = \QOY’ = A Now, In right angled DPMO and DQNO, i. \PMO = \QNO = 90° ii. \POM = \OQN = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ PM = ON and OM = –QN. Then, In right angled DQNO, Sin (270°+A) = OQ NQ = OP –OM = –CosA Cos (270°+A) = OQ ON = OP –PM = SinA Tan (270°+A) = ON QN = PM –OM = –CotA (And so on for the others) vii. Trigonomertical ratios of (360°-A) Let, A revolving line OP makes an angle ‘A’ with OX at the position of OQ. ie. \POX = A \QOX = A \ \QOX (reflex) = 360°-A Draw the perpendiculars, PM⊥OX and QN⊥OX \ \OQN = \QOY’ = A Now, In DPMO and DQNO, i. \PMO = \QNO = 90° ii. \POM = \QON = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ PM = –QN and OM = –ON. Then, In right angled DQON, Sin (360°-A) = OQ QN = OP –PM = –SinA O M Q N Y’ Y X P X’ 270°–A A A A O M N Q Y’ Y X P X’ 360°–A A A A
PRIME Opt. Maths Book - IX 225 Cos (360°-A) = OQ QN = OP OM = CosA TanA (360°-A) = ON QN = OM –PM = –TanA (And so on others can be calculated) viii.Trigonomertical ratios of (360°+A) Let, A revolving line OP makes an angle ‘A’ with OX and again ‘A’ with OX at the position of OQ. ie. \POX = A \QOX = A \ \QOX (reflex) = 360°+A Here, OP and OQ lines at same place (coincide) so far both angles ‘A’ and 360°+A, trigonometric ratios will be same as follows, Sin(360°+A) = OP RM = OP PM = SinA Cos(360°+A) = OQ OM = OP OM = CosA Tan(360°+A) = OM QM = OM PM = TanA (And so on for the others) ix. Trigonomertical ratios of (-A) Let, A revolving line OP makes an angle ‘A’ with OX and again it makes an angle (-A) with OX at the position of OQ. ie. \POX = A (anti-clockwise +ve) \QOX = -A (clockwise -ve) Draw the perpendiculars, PM⊥OX and QN⊥OX \ \OQN = \QOY’ = A Now, In DPMO and DQNO, i. \PMO = \QNO = 90° ii. \POM = \QON = A iii. OP = OQ = radii of same circle. \ DPMO ≅ DQNO (By AAS axiom) \ PM = -QN and OM = -ON. Then, In right angled DQNO, Sin (-A) = OQ QN = O –PM = – SinA Cos(–A) = OQ ON = O OM = CosA Tan(–A) = ON QN = OM –PM = –TanA O M Q Y’ Y X P, Q X’ 360°+A A O M N Q Y’ Y X P X’ A –A
226 PRIME Opt. Maths Book - IX Cosec(–A) = QN OQ = PM OP – = –CosecA Sec(–A) = ON OQ = OM OP = SecA Cot(–A) = QN ON = PM OM – = –CotA. x) Generalization of the angles modified from standard angles as above: Sign rule (CAST RULE) Above derivation of the different forms of angles belonging to the quadrants, we Conclude that: 1st quadrent → All trigonometrical ratios are positive 2nd quadrent → only Sin and Cosec are positive(S) 3rd quadrent → only Tan and Cot are positive (T) 4th quadrent → only Cos and Sec are positive (C) O 1st Quadrent (All) 4th Quadrent (Cos + Sec) S A T C 2nd Quadrent (Sin + Cosec) 3rd Quadrent (Tan + Cot) (180° + A, 270° – A) (90° + A, 180° – A) (360° – A, 270° + A) (360° + A, 90° – A) Y’ Y X’ X 5.9.1 Multiple of 90° rule: Any kind of angles can be derived in the form of (90° × n + A) where n is any constant (Natural number). • If ‘n’ being even, the trigonometrical ratios will not be changed. • If ‘n’ being odd, the trigonometrical ratios will be changed as, Sin → Cos Cot →Tan Cos → Sin Cosec → Sec Tan → Cot Sec → Cosec Examples: Sin(270° + A) = Sin(90° × 3 + A) = Sin (90° × odd + A) It belongs to 4th quadrent. = – CosA Tan(180° + A) = Tan(90° × 2 + A) = Tan(90° × even + A) It belongs to 3rd quadrent. = TanA S A T C Y’ Y X’ X S A T C Y’ Y X’ X
PRIME Opt. Maths Book - IX 227 Cos 120° = Cos(90° × 1 + 30°) = Cos(90° × odd + 30°) It belongs to 2nd quadrent. = –Sin30° =- 2 1 Tan 780° = Tan(90° × 9 – 30°) = Tan (90° × odd – 30°) It belongs to 1st quadrent. = –sin30° = 3 Cos 150° = Cos(90° × 1 + 60°) = Cos(90° × odd + 60°) It belongs to 2nd quadrent. = Cot30° = 3 Cot 150° = Cot(90° × 2 - 30°) = Cot (90° × even + 60°) It belongs to 2nd quadrent. = -Cot 30° =- 3 S A T C Y’ Y X’ X S A T C Y’ Y X’ X S A T C Y’ Y X’ X S A T C Y’ Y X’ X 1. Find the value of Sin(–1140°) Solution: Sin (-1140) = –Sin 1140 = –Sin(90° × 12 + 60°) = –Sin 60°” = 3 2. Prove that: Cos 55° + Sin 75° = Cos 15° + Sin 35° Solution: L.H.S = Cos 55° + Sin 75° = Cos(90° × 1 – 35)+ Sin (90° × 1 – 15°) = Sin 35° + Cos15° = R.H.S Worked out Examples
228 PRIME Opt. Maths Book - IX 3. Prove that: Sin 420° Cos (–390°) + Cos 300. Sin(–330) = 1 Solution: L.H.S = Sin 420° . Cos(–390) + Cos 300° . Sin(–330) = Sin(90° × 4 + 60°) . Cos(90° × 4 + 30) + Cos(90° × 3 + 30°) [\Cos(–q) = Cosq] = Sin 60° . Cos30° + Sin 30° (Cos 60°) = 2 3 × 2 3 + 2 1 × 2 1 = 4 3 + 4 1 = 4 4 = 1 = R.H.S. 4. Prove that: Cos 16 r + Cos 16 5r + Cos 16 11r + Cos 16 15r = 0 Solution: L.H.S = Cos 16 r + Cos 16 5r + Cos 16 11r + Cos 16 15r = Cos 16 r + Cos 16 5r + Cos 16 16r r - 5 a k + Cos 16 16r r- a k = Cos 16 r + Cos 16 5r + Cos 16 5 r r a - k + Cos 16 r r ` j - = Cos 16 r + Cos 16 5r + Cos 90 2 16 5 °× r a - k + Cos 90 2 16 °× – r ` j = Cos 16 r + Cos 16 5r - Cos 16 5r - Cos 16 r = 0 = R.H.S 5. Tan 20 r Tan 20 3r Tan 20 5r Tan 20 7r Tan 20 9r = 1 Solution: L.H.S = Tan 20 r Tan 20 3r Tan 20 5r Tan 20 7r Tan 20 9r = Tan 20 r Tan 20 3r Tan 45°.Tan 20 10r r – 3 a k.Tan 20 10r r – a k = Tan 20 r Tan 20 3r × 1 × Tan 2 20 3 – r r a k.Tan 2 20 – r r ` j = Tan 20 r Tan 20 3r .Tan 90 1 20 3 °× – r a k.Tan 90 1 20 °× – r ` j = Tan 20 r .Tan 20 3r + 20 3r .Cot 20 r = (Tan p 20.Cot p 20) (Tan3p 20.Cot3p 20) = 1 × 1 = 1 = R.H.S
PRIME Opt. Maths Book - IX 229 6. Prove that: Sin(90° + A).Tan(180° – A).Cosec(360° – A) Cos(180° – A).Tan (180° – A).Cosec (360° – A) = –1 Solution: L.H.S = Sin(90° + A).Tan(180° – A).Cosec(360° – A) Cos(180° – A).Tan (180° – A).Cosec (360° – A) = Sin(90° × 1 + A).Tan(90° × 2 – A).Sec(90° × 3 – A) Cos(90° × 2 – A).Tan (90° × 2 + A).Cosec (90° × 4 + A) = CosA(–1TanA).(–CosecA) (–CosA).TanA.CosecA = – CosA.TanA.CosecA CosA.TanA.CosecA = -1 = R.H.S 7. Prove that: Sin2 8 r + Sin2 8 3r + Sin2 8 5r + Sin2 8 7r = 2 Solution: L.H.S = Sin2 p 8 + Sin2 8 3r + Sin2 8 5r + Sin2 8 7r = Sin2 p 8 + Sin2 8 3r + Sin2 8 4r r + a k + Sin2 8 4 3 r r + a k = Sin2 p 8 + Sin2 8 3r + Sin2 2 8 r r ` j + + Sin2 2 8 r r3 a + k = Sin2 p 8 + Sin2 8 3r + Sin2 90 1 8 °× r ` j + + Sin2 90 1 8 3 °× r a + k = Sin2 p 8 + Sin2 8 3r + Cos2 p 8 + Cos2 8 3r = (Sin2 p 8 + Cos2 p 8 ) + (Sin2 8 3r + Cos2 8 3r ) = 1+1 = 2 = R.H.S 8. Find the value of ‘x’ from the followings equation: xSec(180° – A) Cosec(270° + A) – Sin(90° + A). Cos(360° – A) = Cos(270° – A) Sin(180° + A) + x Tan(270° + A). Tan(180° – A) or, x Sec(90° × 2 – A). Cosec(90× 3 + A) –Sin(90° × 1 + A). Tan(90° × 4 – A) = Cos(90× 3 – A). Sin(90° × 2 + A) + x Tan(90° × 3 + A). Tan(90° × 2 – A) or, x (–SecA). (–SecA) – CosA . CosA = (–SinA) (–SinA) + x(–CotA). (–TanA) or, x Sec2 A – Cos2 A = Sin2 A + x × 1 or, x (Sec2 A – 1) = Cos2 A + Sin2 A or, x = Tan A 1 2 \ x = Cot2 A.
230 PRIME Opt. Maths Book - IX 9. Find the value of ‘A’ from SinA = Cos2A. Solution: SinA = Cos2A or, SinA = Sin(90° – 2A) or, A = 90° – 2A or, 3A = 90° or, A = 3 90° 30° \ A = 30° Exercise 5.9 1. i) Show the sign rule in a quadrant. ii) Write down the conditional of trigonometric ratio changed or not according to CAST rule in a quadrant. iii) Simplify Sin(270° – q) by showing in quadrant. iv) Simplify Tan(180° – q) by showing in quadrant. v) Simplify Cos(540° – q) by showing in quadrant. 2. Find the value of the following: i) Sin 240° ii) Cos(-780°) iii) Tan 960° iv) Cot(-1200°) v) Cosec 1830° 3. Find the value of the followings: i) Sin2 120° + Cos2 150° – Tan2 135° ii) 4 3 Tan2 150°- 2 1 Cos 180° + Tan 135° iii) 2Cos2 135° + Sin2 150° + 2 1 Cos 180° + Tan 135° iv) Sin2 135° Cos2 120° - Sin2 120°Cos2 135° v) Sin2 4 r + Sin2 4 3r + Sin2 4 5r + Sin2 4 7r 4. Find the value of the followings. i) Cos2 150° + Sin2 120° + Sin2 150° + Cos2 120° ii) Sin 300°.Cos(-330°) + Cos 420°.Sin(-390°) iii) Cos 330° Sin 300° – Sin 330° Cos 660° 2Sin 750° iv) Cos 510° Sin 600° – Cos 420° Sin (–630°) v) Sin570°Cos420° + Cos750°Sin600° 5. Prove that the following: i) Tan 32° + Cos 53° – Cosec 70° = Sin 37° + Cot 58° – Sec 20° ii) Tan 9° Tan 27° Tan 45° Tan 63° Tan 81° = 1
PRIME Opt. Maths Book - IX 231 iii) Cos 112° + Sin 74° + Cos 68° – Sin 106° = 0 iv) Cos 306° + Cos 234° + Cos 162° + Cos 18° = 0 v) Cos 20° + Cos 40° + Cos 140° + Cos 160° = 0 vi) Cot 9° Tan 27° – Tan 81° Cot 63° = 0 6. Prove that the followings: i) ( ) ( ) Tan Sec 270 360 ° °– i i + × ( ) ( ) Cot A Cos A 360 180 °– °– × ( ) ( ) Cos A Sin A 270 90 ° – ° + = TanA ii) ( ) ( ) Tan Sin 180 180 ° °– i i + . ( ) ( ) Sin Cos 360 360 °– ° i + i . ( ) ( ) Tan Cot 90° – i i + = – Sinq iii) ( ) ( ) Tan A Sin A 180 180 °– ° + . ( ) ( ) Tan A Cot A 90 270 °– °– . ( ) ( ) Sin A Cos A 360 – °– = SinA iv) Tan(180° – q).Cot(90° – q).Cos(360° + q) Cot(270° – q).Cot(180° – q).Sin (–q) = –Tanq v) Sin(90°+q).Cos(180°–q).Tan(360°–q) = Sin2 (90°–q).Cot(270°–q) 7. Prove that the followings: i) Cos102° + Cos126° + Sin36° + Sin12° = 0. ii) Cos p 8 + Cos 8 3r + Cos 8 5r + Cos 8 7r = 0. iii) Sin 13 2r + Sin 13 3r – Sin 13 10r – Sin 13 11r = 0. iv) Cos 16 3r + Cos 16 5r + Cos 16 11r + Cos 16 13r = 0. v) Cos 24° + cos 48° + Cos 132° + Cos 156° = 0 8. Prove that the following: i) Sin2 8 r + Sin2 8 3r + Sin2 8 5r + Sin2 8 7r = 2 ii) Cos2 16 r + Cos2 16 3r + Cos2 16 5r + Cos2 16 7r = 2 iii) Tan12° Tan36° Tan84° Tan108° = 1 iv) Cot 20 r Cot 20 3r Cot 20 5r Cot 20 7r Cot 20 9r = 1. v) Sin2 32 3r + Sin2 32 5r + Sin2 32 11r + Sin2 32 13r = 2. 9. Find the value of ‘x’ from the followings: i) x Sin p 4 Cos p 4 . Tan p 4 – Tan2 p 4 + Cos2 p 4 = 0 ii) x Cotq. Tan(270° + q) = Tan (90° + q). Cot (180 – q) + x Sec(90° + q). Cosec(180° – q) iii) Tan(270° + A) Sin(180° + A). Cosec(270° – A) = x Sin (180° + A) – Cos2 (90° + A) Tan2 (450° + A) iv) 2Cot120°. Tan135° Cos180° – x sin120° Cos180° = xTan150° v) x Cos(90° + q). Cos(90° – q) + Tan(180° – q). Cot(90° + q) = Sin(180° – q). Sin(360° + q)
232 PRIME Opt. Maths Book - IX 10. Find the value of the angle ‘A’ from the followings: i) 3 TanA + 1 = 0 ii) 4CoaA + 2 = 0 iii) SinA + CosA = 0 iv) Sin2A + SinA = 0 v) Cot3A + TanA = 0 11. PRIME more creative questions: i) Prove that : 1 + Cos 7 2r Cos 7 5r = Sin 7 2r Sin 7 5r . ii) If A, B and C are the angles of a triangle, prove that: Tan A B 2 + a k = Cot C 2 . iii) If A, B, C and D are the angles of a quadrilatral, prove that: Cos A B 2 + a k + Cos C D 2 + a k = 0. iv) If P, Q, R and S are the angles of a quadrilateral, prove that CosA + CosB + CosC + CosD = 0. v) If x Cos(180° + q). Cot(270° + q) = Sin(90° + q) + Cosec(270° – q), find the value of ‘x’. Answer 1. Show to your teacher. 2. i) 2 3 c m – ii) 2 1 iii) 3 iv) 3 1 v) 2 3. i) 2 1 ii) 4 1– iii) 4 1– iv) 4 1– v) 2 4. i) 2 ii) –1 iii) 2 1 iv) 1 4 1 v) – 1 9. i) 3 ii) Cot2 q iii) SinA iv) 5 4 v) Tan2 q 10. i) 150° ii) 120° iii) 135° iv) 180° v) 45° 11. v) –Tanq
PRIME Opt. Maths Book - IX 233 5.10 Compound Angle (A + B) and (A – B) are taken as the compound angle of the angles ‘A’ and ‘B’ . Here, we discussed about the trigonometrical ratios of such compound angles A + B, A – B, A + B + C, A – B – C etc. 5.10.1 Trigonometrical ratios of (A + B) Let, a revolving line OY makes and angle A with OX and OZ makes an angle B with OY. i.e. ∠YOX = A, ∠ZOY = B ∠ Taking a point p on OZ, Draw the perpendiculars PM⊥OX, PQ⊥OY, QN⊥OX & QR⊥PM where, ∠RQO = ∠YOX = A ∠PQR = 90° – A ∠RPQ = 90° – ∠PQR = 90° – (90° – A) = A. Then, In right angled ∆PMO, ∠POM = A + B i) Sin(A + B) = OP PM = OP PR + RM = OP PR OP QN + = × PQ PR OP PQ + × OQ QN OP OQ = CosA × SinB + SinA × CosB ∴ Sin(A+B) = SinA.CosB + CosA.SinB ii) Cos(A+B) = OP OM = OP ON – MN = – OP ON OP RQ = × OR ON OP OR – × PQ RQ OP PQ ∴ Cos(A+B) = CosA.CosB – SinA.SinB iii) Tan(A+B) = ( ) ( ) Cos A B Sin A B + + = SinA.CosB + CosA.SinB CosA.CosB – SinA.SinB Dividing each terms by CosA.CosB = SinA.CosB + CosA.SinB CosA.CosB CosA.CosB – SinA.SinB CosA.CosB O X’ M R N A A B Q P Z Y
234 PRIME Opt. Maths Book - IX Tan(A+B)= TanA.TanB TanA TanB 1 – + Cot(A+B) = ( ) ( ) Sin A B Cos A B + + = CosA.CosB +SinA.SinB SinA.SinB SinA.CosB + CosA.CosB SinA.SinB \ Cot(A+B) = . CotB CotA CotA CotB – 1 + 5.10.2 Trigonometrical ratios of (A – B) Sin(A – B) = SinACosB – CosASinB Cos(A – B) = CosACosB + SinASinB Tan(A – B) = TanA – TanB 1 + TanA.TanB Cot(A – B) = CotA.CotB + 1 CotB – CotA 5.10.3 Trigonometrical ratios of A + B + C i) Sin(A + B + C) = Sin{(A + B) + C} = Sin(A + B)CosC + Cos(A + B)SinC = (SinACosB + CosASinB) CosC + (CosACosB – SinASinB)SinC = SinACosBCosC + CosASinBCosC + CosACosBSinC – SinASinBSinC ii) Cos(A + B + C) = Cos{(A + B) + C} = Cos(A + B)CosC – Sin(A + B)SinC = (CosACosB – SinASinB)CosC – (SinACosB + CosASinB)SinC = CosACosBCosC – SinASinBCosC – SinACosBSinC – CosASinBSinC iii) Tan(A + B + C) = Tan {(A + B) + C} = Tan(A + B) + TanC 1 – Tan(A + B).TanC = TanA + TanB 1 – TanA.TanB TanA + TanB 1 – TanA.TanB + TanC 1 – TanC = TanA + Tan B + TanC – TanA.Tan B.TanC 1 – TanA.TanB – TanB.TanC – TanC.TanA
PRIME Opt. Maths Book - IX 235 iv) Cot(A + B + C) = Cot{(A + B) + C} = Cot(A + B).CotC – 1 CotC + Cot(A + B) = CotA.CotB – 1 CotB + CotA CotA.CotB – 1 CotB + CotA × CotC – 1 CotC + = CotA.CotB.CotC. – CotA – CotB – CotC CotA.CotB + CotB.CotC + CotC.CotA – 1 1. Find the value of Sin15° without using calculator. Solution : Sin15° = Sin(45° – 30°) = Sin45°Cos30° – Cos45°Sin30° = 2 1 × 2 3 – 2 1 2 1 × = 2 2 3 1– 2. If SinA = 5 3 , CosB = 13 12 , find the value of Sin(A + B). Solution : SinA = 5 3 , CosB = 13 12 \ CosA = 1 – Sin A2 , SinB = 1 – Cos B2 = 1 5 3 – 2 a k = 1 13 11 – 2 a k = 5 4 = 13 5 Then, Sin(A+B) = SinA.CosB + CosA.SinB = × 5 3 13 12 + × 5 4 13 5 = 65 36 + 20 = 65 56 3. If A + B = 45°, prove that (CotA – 1)(CotB – 1) = 2 Solution: Taking, A + B = 45° or, Cot(A + B) = Cot45° or, . CotB CotA CotA CotB – 1 + = 1 or, CotA.CotB – 1 = CotB + CotA or, CotA.CotB – CotB – CotA = 1 Worked out Examples
236 PRIME Opt. Maths Book - IX Adding ‘1’ on both sides, or, CotA.CotB – CotB – CotA + 1 = 1 + 1 or, CotB(CotA – 1) –1(CotA – 1) = 2 or, (CotA – 1) (CotB – 1) = 2 L.H.S. = R.H.S. Proved 4. Prove that the followings. i) Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A ii) Cot20°.Cot50° + Cot50°.Cot110° + Cot110°.Cot20° = 1 Solution: i) We have, 7A = 4A + 3A or, Tan(7A) = Tan(4A + 3A) or, Tan7A = Tan4A + Tan3A 1 – Tan4A.Tan3A or, Tan4A + Tan3A = Tan7A – Tan7A.Tan4A.Tan3A or, Tan7A.Tan4A.Tan3A = Tan7A – Tan4A – Tan3A \ Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A. L.H.S. = R.H.S. ii) We have, 20° + 50° + 110° = 180° or, 20° + 50° = 180° – 110° or, Cot(20 + 50) = Cot(180° – 110°) or, Cot20°. Cot50° – 1 Cot50° + Cot20° = – Cot110° or, Cot20°.Cot50° – 1 = – Cot50°.Cot110° – Cot110°.Cot20° or, Cot20°.Cot50° + Cot50°.Cot110° + Cot110°.Cot20° = 1 L.H.S. = R.H.S. 5. If an angle A is divided into two parts a and b that bTana = aTanb, prove that Sin(a – b) = a b a b – + SinA. Solution : Here, bTana = aTanb or, b = Tan aTan a b = Sin Cos aSin Cos a b b a Then, R.H.S. = a b a b SinA – + = SinA a – a + Sinb.Cosa Sina.Cosb aSinb.Cosa Sina.Cosb = a(Sina.Cosb – Sinb.Cosa) a(Sina.Cosb + Sinb.Cosa) SinA = Sin(a – b) Sin(a + b) Sin(a + b) Sin(a – b) = L.H.S. proved
PRIME Opt. Maths Book - IX 237 Exercise 5.10 1. i) Write down the formula of Sin(A + B). ii) Write down the formula of Cos(a – b). iii) Write down the formula of Tan(A + B). iv) Prove that Cot(A + B) = CotA.CotB – 1 CotB + CotA v) Prove that Tan(a + b) = Tana – Tanb 1 + Tana.Tanb 2. i) If SinA = 5 4 , SinB = 13 5 , find the value of Cos(A + B). ii) If CosA = 2 1 , CosB = 2 3 , find the value of Sin(A – B). iii) If 3TanA = 4, and TanB = 12 5 find the value of Tan(A + B) iv) If Cota = 3 and Cotb = 2, find the value of Tan(a – b). v) If Tan(A + B) = 1 and Tan A = 2 1 , find the value of TanB. 3. Find the value of the followings with out using calculator. a) i) Sin 75° ii) Cos 15° iii) Tan 105° iv) Cos 285° v) Sin 555° b) i) Tan 15° + Cot 15° ii) Sin 15° + Sin 75° iii) Cos 105° + Cos 165° iv) Sin 165° + Sin 75° v) Cos 195° + Sin 285° 4. Prove that the following using given condition. a) i) If A + B = 45°, prove that : CotA.CotB – CotA – CotB = 1 ii) If A + B = 45°, prove that : (1 + TanA)(1 + TanB) = 2 iii) If a + b = 225°, Prove that : (Cota – 1)(Cotb – 1) = 2 iv) If A + B = p 4 , prove that : TanA(TanB + 1) = 1 – TanB v) If a + b = p 4 , prove that : ( ) ( ) Cot Cot Cot 1 – 1 a b a + = 1 b) i) If A + B + C = 180°, prove that : TanA + TanB + TanC = TanA.TanB.TanC ii) If a + b + g = p, prove that : Cota.Cotb + Cotb.Cotg + Cotg.Cota = 1 iii) If A + B + C = p, prove that : Tan A 2 .Tan B 2 + Tan B 2 .Tan C 2 + Tan. C 2 Tan A 2 = 1 iv) If A + B + C = p and CosA = CosB.CosC, prove that : TanA = Tan B + TanC. v) If A + B + C = p and CosA = CosB + CosC, Prove that : CotB.CotC = 2 1 c) i) If TanA = 4 3 , TanB = 7 1 , prove that A + B = 45°. ii) If TanA = m m + 1 , Tan B = 2 1 m 1 + , prove that A + B = 4 c r .
238 PRIME Opt. Maths Book - IX iii) If Cota = 11 and Cotb = 5 6 , prove that a + b = 4 c r . iv) If Sina = 10 1 & Cosb = 5 2 , show that a + b = 45°. v) If Cosq = 5 4 & Cosf = 5 2 7 , show that q – f = 4 c r . 5. Prove that the followings: a) i) Sin(A + 45°) = 2 1 (SinA + CosA) ii) Cos(A – 45°) = 2 1 (CosA – SinA) iii) Sin(A + 45°) – Sin(A – 45°) = 2 CosA. iv) Sin(A + 30°) = 2 1 ( 3 SinA + CosA) v) Cos( p 3 + q) + Cos( p 3 – q) = Cosq. b) i) Tan (45° + a) = Cos Sin Cos Sin a a – a a + ii) Cot ( p 4 – q) = Cos Sin Cos Sin i i – i i + iii) Cos(60° + A) + Cos(60° – A) = 0 iv) Sin( 3 2r + A) + Sin( 3 2r – A) – 3 CosA = 0 v) Tan(45° + A) – Tan(45° – A) = Tan Tan 1 4 – 2 i i c) i) Sin(A + B) + Sin(A – B) = 2SinA.CosB ii) Cos(A + B) + Cos(A – B) = 2CosA.cosB iii) Sin(A + B) – Sin(A – B) = 2CosA.SinB iv) Cos(A – B) – Cos(A + B) = 2SinA.SinB v) Tan(A + B) – Tan(A – B) = 2Sec2 A.TanB 1 – Tan2 A.Tan2 B d) i) 1 – Tan13°Tan32° = Tan13° + Tan32° ii) (1 + Tan22°) (1 + Tan23°) = 2 iii) (Cot20° – 1) (Cot25° – 1) = 2 iv) Cot 100° + Cot 35° + Cot 100°.Cot35° = 1 v) (Tan55° – 1°) (Tan80° – 1) = 2 e) i) Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A ii) Tan70° – Tan50° – Tan20° = Tan70°.Tan50°.Tan20° iii) Cot10°.Cot15° – Cot15°.Cot25° – Cot25°Cot.10° = 1 iv) CotA.Cot2A – CotA.Cot3A – Cot3A.CotA = 1 v) Sin25°.Cos20° + Cos25°.Sin20° = 2 1 f) i) Tan20° + Tan75° + Tan85° = Tan20°.Tan75°.Tan85° ii) Cot10°.Cot70° + Cot70°Cot100° + Cot100°Cot10° = 1
PRIME Opt. Maths Book - IX 239 iii) Cos10° – Sin10° Cos10° + Sin10° = Tan 35° iv) Cos14° + Sin14° Cos14° – Sin14° = Cot31° v) Cos Sin Cos Sin 8 8 8 8 ° – ° ° ° + = Tan53° g) i) . ( ) CosA CosB Sin A – B = TanA – TanB ii) Sin2 (45° + q) + Sin2 (45° – q) = 1 iii) Cot(45° + A).Cot(45° + A) = 1 iv) Tan(A+B).Tan(A – B) = Cos B – Sin A Sin A Sin B 2 2 2 2 + v) Sec A Sin A 2 3 + Co sec A Cos A 2 3 = Sin5A h) i) Tan70° = 2Tan50° + Tan20° ii) Tan50° = 2Tan10° + Tan40° iii) Cos18° – Sin18° = 2 Sin27° iv) Cos10° + Sin10° = 2 Sin55° v) Tan(a + b) – Tana 1 + Tan(a + b).Tana = Tanb 6. PRIME more creative questions: a) i) Prove that : Sin(A + B + C) = CosA.CosB.CosC(TanA + TanB + TanC – TanA.TanB.TanC) ii) Prove that : SinA.Sin(B – C) + SinB.Sin(C – A) + SinCSin(A – B) = 0 iii) Prove that : Tan(a – b) + Tan(b – g) + Tan(g – a) = Tan(a – b).Tan(b – g). Tan(g – a) iv) If A + B + C = p and CosA = CosB.CosC, prove that : TanA = TanB + TanC v) If A + B + C = p and SinA = SinB.SinC, prove that : CotB + CotC = 1 b) i) Prove that : Sin2 A + Sin2 B + 2SinASinB.Cos(A + B) = Sin2 (A + B) ii) Prove that : . ( ) CosA CosB Sin A – B + . ( ) CosB CosC Sin B – C + . ( ) CosC CosA Sin C – A = 0 iii) An angle q is divided into A and B such that TanA.TanB = m:n, prove that : Sin(A – B) = m n m n – + ` j.Sinq. iv) If Tana = kTanb, prove that : ( ) ( ) Sin Sin a b– a b + = k k 1 1 – + v) If 2TanA + CotA = TanB, prove that: 2Tan(B – A) = CotA. 7. Project work Collects the formula used in trigonometry in a chart paper and present into the classroom and paste at the project board. Answer 1. Show to your teacher. 2. i) 65 16 ii) 2 1 iii) 1 16 15 iv) 7 1– v) 3 1 3. i) 4 6 2 + ii) 4 6 2 + iii) –( ) 2 3 + iv) 4 6 2 – v) – 4 2 6
240 PRIME Opt. Maths Book - IX Co-ordinate Geometry Unit Test - 1 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. What is radian angle? Write down its value in degrees. 2. a. Express 35° 42’ 54” into centesimal measurement. b. Prove that: Cosp 8 + Cos 8 3π + Cos 8 5π + Cos 8 7π = 0 c. If m SinA = n, find sec2 A – Tan2 A. 3. a. Prove that: CotA TanA 1 – + TanA CotA 1 – = SecA.CosecA + 1 b. If x Cotq. Tan(90° + q) = Tan(270° + q). Cot(180° – q) – xSec(90° – q). Cosec(360° – q), find the value of x. 4. If the angles of a triangle in degrees, grades and radians respectively are in the ratio 288: 280: p, find the angles of the triangle in degrees. Unit Test - 2 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. Write down the formula to find central angle of a circle with the usual meaning of the symbols used. 2. a. Find the radius of a circle forms a central angles of 60° with an arc of 8.8 cm. b. Prove that: Sin Cos 1 60 1 30 ° – ° + = 7 – 4 3 c. Prove that: Sin A 1 – Cos A 4 4 = 2Cosec2 A – 1 3. a. If 5Cosq + 12Sinq = 13, find Cotq. b. Prove that: Sin2p 8 + Sin2 8 3π + Sin2 8 5π + Sin2 8 7π = 2. 4. Prove that: (3 – 4Sin2 q) (1 – 3Tan2 q) = (3 – Tan2 q) (4Cos2 q – 3)
PRIME Opt. Maths Book - IX 241 Unit 6 Vector Geometry Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions 1 2 – 1 4 10 12 Weight 1 4 – 5 K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Student are able to denote the vector and scalar with examples. • Students are able to identify the types of vectors. • Students are able to find the operation of vectors like addition, subtraction multiplication by scholar, unit vector etc. • Students are able to find magnitude and direction of vectors. • Students are able to the simple geometry involving vector. Materials Required: • Chart paper. • Graph paper. • Flash card. • Sample of position vector of a point in chart paper. • Chart of types of vector and operation on vector. • Chart of vector to find magnitude and direction.
242 PRIME Opt. Maths Book - IX 6.1 Vector and Scalar Out of different physical quantities in our daily life, some of them have magnitude only while some of the others have magnitude as well as direction. The physical quantities which have only magnitude are the scalar quantities and which have magnitude as well as direction are called vector quantities. Vector: The physical quantity which has magnitude as well as direction is called vector. Eg : displacement, velocity, force, acceleration, etc. Scalar: The physical quantity which has only magnitude is called scalar quantity. Eg : distance, area, speed, density, mass, length etc. 6.1.1 Representation of vector & Scalar: • Vector quantity is represented by using a directed line segment with arrow. B A , AB , a etc. • Scalar quantity is represented by a line segment only. A B, AB, a etc. • The line segment having speci�ic direction is called directed line segment. P Q • Vector is represented by using two components ‘x’ and ‘y’ as AB = x Component y Component - - e o Where, for a point A(x, y), components of OA are (Projection of OA on x-axis) (Projection of OA on y-axis) x – component = OM = x y – component = AM = y ∴ OA = x Comp y Comp – – d n= x y d n A(x, y) X Y O M
PRIME Opt. Maths Book - IX 243 • If A(x1 , y1 ) and B(x2 , y2 ) are given vector AB is calculated as shown in diagram. A B (x2 , y2 ) (x1 , y1 ) P Y X M N M’ N’ O x y Where, Here, AP = x – Comp. = MN = ON – OM = x2 – x1 (projection of AB on x-axis) BP = y – Comp. = M’N’ = BN – PN = BN – AM = y2 – y1 (projection of AB on y-axis) \ AB = . . x comp y comp - - d n = x x y y – – 2 1 2 1 d n • Study the following graph having directed line segment. Here, From graph we get, AB = . . x comp y comp - - d n = 3 2 d n PQ = . . x comp y comp - - d n = 3 –4 d n RS = . . x comp y comp - - d n = 4 4 – – d n MN = . . x comp y comp - - d n = 3 5 – d n 6.1.2 Position Vector of a point: In the vector OA , the initial point is the origin and final point is A(4, 3). Where, x - component = OM = 4 y - component = AM = 3 And position vector of a point A = OA = 4 3 d n Y X A(4, 3) O M 3 4 A P Q R S N M B The position vector of a point P(x, y) is the vector OP where ‘O’ is the origin and x y OP = d n.
244 PRIME Opt. Maths Book - IX 6.1.3 Types of vector i) Column vector: The vector whose components are written in a column is called column vector. In the diagram, AB = . . x comp y comp - - d n = 4 1 4 2 – – d n = 3 2 d n ii) Row vector : The vector whose components are written in a row is called row vector. In the given diagram, AB = (x - comp. y - comp) = (4 6) iii) Zero vector (Null) The vector having magnitude zero is called zero vector whose components are also zero. i.e. AA = 0 0 d n, BB = 0 0 d n The direction of zero vector is not fixed. iv) Equal vectors: Any two vector having same direction as well as equal magnitude are called equal vectors. Eg : a A B a P Q Here, AB = = PQ a v) Negative vectors: Any two vectors having equal magnitude but in opposite direction are called negative vectors. Eg: a A B A B – a Here, AB = a BA = a (in opposite direction) i.e. AB = a = – (– a ) = –BA \ AB & BA are negative vector i.e. AB = – BA Y X A(1, 2) B(4, 4) O M 3 N C Y X A(1, 2) B(5, 7) O M 4 N
PRIME Opt. Maths Book - IX 245 vi) Co-intial vectors: The vectors having same initial point are called co-initial vectors. Here, The vectors OA and OB have the same initial point ‘O’. Hence, they are co-initial vectors. But OA ≠ OB vii) Like vectors Any two vectors having same direction are called like vectors. The like vectors may be different in their magnitude. viii)Unlike vectors Any two vectors having opposite direction are called unlike vectors. The unlike vectors may be different in their magnitude. ix) Parallel vectors Any two vectors which are in the form of multiplication with a constant number to each other are called parallel vectors. i.e. a = mb and b = na . They are like as well as unlike. Example : If p = 2 1 d n and q = 6 3 d n, Here, q = 6 3 d n = 3 2 1 d n = 3p which is same as a = mb for a ||b \ p ||q is proved 1. If a point P(3, 4) is directed by a line segment from the origin, find OP and its components. Solution : Initial point is 0(0, 0) Final point is P(3, 4) \ x-component = x2 – x1 = 3 – 0 = 3 y - component = y2 – y1 = 4 – 0 = 4 Then, OP = x comp y comp - - d n = 3 4 c m Worked out Examples B O A