Prove that tan^{- 1} 63/16 = sin^{- 1} 5/13 + cos^{- 1} 3/5

Solution:

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios. 

Here the basic trigonometric function of Sin θ = y, can be changed to θ = sin^-1 y

Let sin^{- 1} 5/13 = x

⇒ sin x = 5/13

Then,

cos x = √1 - (5/13)² = 12/13

Therefore,

tan x = 5/12

x = tan^{- 1} 5/12

sin^{- 1} 5/13 = tan^{- 1} 5/12 ....(1)

Now, let cos^{- 1} 3/5 = y

⇒ cos y = 3/5

Then,

sin y = √1 - (3/5)²

= √16/25

= 4/5

Therefore,

tan y = 4/3

y = tan^{- 1} 4/3

cos^{- 1} 3/5 = tan^{- 1} 4/3 ....(2)

Thus, by using (1) and (2)

LHS = sin^{- 1} 5/12 + cos^{- 1} 3/5

= tan^{- 1} 5/12 + tan^{- 1} 4/3

= tan^{- 1} [(5/12 + 4/3)/(1 - 5/12.4/3)]

= tan^{- 1} [((32 + 45)/60/(60 - 24)/60]

= tan^{- 1} 63/16

= LHS

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 7

Prove that tan^{- 1} 63/16 = sin^{- 1} 5/13 + cos^{- 1} 3/5

Summary:

Hence we have proved by using inverse trigonometric functions that tan^{- 1} 63/16 = sin^{- 1} 5/13 + cos^{- 1} 3/5