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RD Chapter 6 Trigonometric Identities Ex 6.1 Solutions

Question - 51 : -

Answer - 51 : -


Question - 52 : -

Answer - 52 : -


Question - 53 : - sin2 A cos2 B– cos2 A sin2 B = sin2 A – sin2 B.

Answer - 53 : -

= sin2 A(1 – sin2 B) – (1 – sin2 A) sin2 B
= sin2 A – sin2 A sin2 B – sin2 B+ sin2 A sin2 B
= sin2 A – sin2 B
Hence, L.H.S. = R.H.S.

Question - 54 : -

Answer - 54 : -


Question - 55 : - cot2 A cosec2 B– cot2 B cosec2 A = cot2 A –cot2 B

Answer - 55 : -


Question - 56 : - tan2 A sec2 B– sec2 A tan2 B = tan2 A – tan2 B

Answer - 56 : -


Question - 57 : - Prove the following identities: (58-75)
If x = a sec θ + b tan θ and y = a tan θ + bsec θ, prove that x1 – y2 = a2 –b1. [C.B.S.E. 2001, 20O2C]

Answer - 57 : -

x – a sec θ + b tan θ
y = a tan θ + b sec θ
Squaring and subtracting, we get
x2-y2 = {a sec θ + b tan θ)2 – (atan θ + b sec θ)2
= (a2 sec2 θ + b2 tan2  θ+ 2ab sec θ x tan θ) – (a2 tan2  θ + b2 sec2  θ+ 2ab tan θ sec θ)
= a2 sec2 θ + b tan2  θ +lab tan θ sec θ – a2 tan2  θ – b2 sec2  θ– 2ab sec θ tan θ
= a2 (sec2 θ – tan2  θ) +b2 (tan2  θ – sec2  θ)
= a2 (sec2 θ – tan2  θ) –b2 (sec2  θ – tan2  θ)
=  a2 x 1-b2 x 1 =a2-b2 =R.H.S.

Question - 58 : - If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ –3 cos θ = ±3

Answer - 58 : -


Question - 59 : - If cosec θ + cot θ = mand cosec θ – cot θ =n,prove that mn= 1

Answer - 59 : -


Question - 60 : -

Answer - 60 : -


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