# Prove 2 tan^{- 1} 1/2 + tan^{- 1} 1/7 = tan^{- 1} 31/17

**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

Since we know that

Using trigonometric identity

2 tan^{- 1} x = tan^{- 1} (2x / (1 - x^{2}))

and tan^{- 1} x + tan^{- 1} y

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

Now,

On substituting the value of x and y, we get

LHS = 2 tan^{- 1} 1/2 + tan^{- 1} 1/7

= tan^{- 1} [(2 × 1/2)/(1 - (1/2)^{2})] + tan^{- 1} 1/7

= tan^{- 1} (4/3) + tan^{- 1} 1/7

= tan^{- 1} [(4/3.1/7) + (1 - 4/3.1/7)]

= tan^{- 1} [(28 + 3)/21 + 21/(21 - 4)/21]

= tan^{- 1} (31/17)

= RHS

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 4

## Prove 2 tan^{- 1} 1/2 + tan^{- 1} 1/7 = tan^{- 1} 31/17

**Summary:**

Hence we have proved that 2 tan^{- 1} 1/2 + tan^{- 1} 1/7 = tan^{- 1} 31/17. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios