# Prove that sin^{- 1 }8/17 + sin^{- 1} 3/5 = tan^{- 1} 77/36

**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} 8/17 = x

⇒ sin x = 8/17

Then,

cos x = √1 - (8/17)²

= √225/289 = 15/17

Therefore,

tan x = 8/15

x = tan^{- 1} 8/15

sin^{- 1} 8/17 = tan^{- 1} 8/15 ....(1)

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

⇒ sin y = 3/5

Then,

cos y = √1 - (3/5)²

= √16/25 = 4/5

Therefore,

tan y = 3/4

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

Thus, by using (1) and (2)

LHS = sin^{- 1 }8/17 + sin^{- 1} 3/5

= tan^{- 1 }8/15 + tan (3/4)

= tan^{- 1 }[8/15 + (3/4)]

= tan^{- 1 }[(8/15 + 3/4)/(1 - 8/15.3/4)]

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

= tan^{- 1} 77/36

= RHS

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 4

## Prove that sin^{- 1 }8/17 + sin^{- 1} 3/5 = tan^{- 1} 77/36

**Summary:**

Hence we have proved by using inverse trigonometric functions that sin^{- 1 }8/17 + sin^{- 1} 3/5 = tan^{- 1} 77/36