# Solve tan^{-1} (x/y) - tan^{- 1} (x - y)/(x + y) is equal to

(A) π/2 (B) π/3 (C) π/4 (D) - 3π/4

**Solution:**

Inverse trigonometric functions as a topic of learning are closely related to the basic trigonometric ratios.

The domain (θ value) and the range(answer) of the trigonometric ratio are changed to the range and domain of the inverse trigonometric function.

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

tan^{- 1} (x/y) - tan^{-1} (x - y)/(x + y)

= [(x/y - (x - y)/(x + y))]/[1 - (x/y)(x - y)/(x + y))]

= tan^{- 1} {[x (x + y) - y (x - y)] / y (x + y)} / {[y (x + y) + x (x - y)] / y (x + y)}

= tan^{- 1} [(x^{2} + xy - xy + y^{2})/(xy + y^{2} + x^{2} - xy)]

= tan^{- 1} (1)

= tan^{- 1} (tan π / 4)

= π / 4

Thus, the correct option is C

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 17

## Solve tan^{-1} (x/y) - tan^{- 1} (x - y)/(x + y) is equal to (A) π/2 (B) π/3 (C) π/4 (D) - 3π/4

**Summary:**

On solving tan^{-1} (x/y) - tan^{- 1} (x - y)/(x + y), we got the value as π/4. Hence correct option is C

visual curriculum