Solve tan^{- 1} (1 - x)/(1 + x) = 1/2 tan^{- 1} x, ( x > 0)

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,

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

Hence,

⇒ tan^{- 1} (1 - x)/(1 + x)

= 1/2 tan^{- 1} x

⇒ tan^{- 1} 1 - tan^{- 1} x

= 1/2 tan^{- 1} x

⇒ π/4 = 3/2 tan^{- 1} x

⇒ tan^-1 x = π / 6

⇒ x = tan π / 6

⇒ x = 1 / √3

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 14

Solve tan^{- 1} (1 - x)/(1 + x) = 1/2 tan^{- 1} x, ( x > 0)

Summary:

For the given function : tan^{- 1} (1 - x)/(1 + x) = 1/2 tan^{- 1} x, ( x > 0), the value of x is 1 / √3. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios