# Prove that cos^{- 1} 4/5 + cos^{- 1} 12/13 = cos^{- 1} 33/65

**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, cos^{- 1} 4/5 = x

⇒ cos x = 4/5

Then,

sin x = √1 - (4/5)² = 3/5

Therefore,

tan x = 3/4

x = tan^{- 1} 3/4

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

Now, let cos^{- 1} 12/13 = y

⇒ cos y = 12/13

Then,

sin y = 5/13

Therefore,

tan y = 5/12

y = tan^{- 1} 5/12

cos^{- 1} 12/13 = tan^{- 1} 5/12 ....(2)

Thus, by using (1) and (2)

cos^{- 1} 4/5 + cos^{- 1} 12/13

= tan^{- 1} 3/4 + tan^{- 1} 5/12

= tan^{- 1} [(3/4.5/12)/(1 - 3/4.5/12)]

= tan^{- 1} [56/33] ....(3)

Now, let cos^{- 1} 33/65 = z

⇒ cos z = 33/65

Then,

sin z = √1 - (33/65)²

= 56/65

Therefore,

tan z = 33/56

z = tan^{- 1} 56/33

cos^{- 1} 33/65 = tan^{- 1} 56/33 ....(4)

Thus, by using (3) and (4)

cos^{- 1} 4/5 + cos^{- 1} 12/13 = cos^{- 1} 33/65

Hence proved

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 5

## Prove that cos^{- 1} 4/5 + cos^{- 1} 12/13 = cos^{- 1} 33/65

**Summary:**

Hence we have proved by using inverse trigonometric functions that cos^{- 1} 4/5 + cos^{- 1} 12/13 = cos^{- 1} 33/65