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Find the Value of `Tan 1/2 [Sin^(-1) (2x)/(1+ X^2) + Cos^(-1) (1-y^2)/(1+Y^2)], |X| < 1, Y> 0 and Xy < 1` - Mathematics

Find the value of `tan  1/2 [sin^(-1)  (2x)/(1+ x^2) + cos^(-1)  (1-y^2)/(1+y^2)], |x| < 1, y> 0  and xy < 1`

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Solution

Let x = tan θ. Then, θ = tan−1 x.

`:. sin^(-1)  (2x)/(1+x^2 ) = sin^(-1)  ((2tan theta)/(1 + tan^2 theta)) = sin^(-1) (sin 2 theta) = 2theta = 2 tan^(-1) x`

Let y = tan Φ. Then, Φ = tan−1 y.

`:. cos^(-1)  (1 - y^2)/(1+ y^2) = cos^(-1) ((1 - tan^2 phi)/(1+tan^2 phi)) =    cos^(-1)(cos 2phi) = 2phi = 2 tan^(-1) y`

`:. tan  1/2 [sin^(-1)  "2x"/(1+x^2) + cos^(-1)  (1-y^2)/(1+y^2)]`

`= tan  1/2 [2tan^(-1) x + 2tan^(-1) y]`

`= tan[tan^(-1) x + tan^(-1) y]`

`= tan[tan^(-1) ((x+y)/(1-xy))]`

`= (x+y)/(1-xy)`

 

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APPEARS IN

NCERT Class 12 Maths
Chapter 2 Inverse Trigonometric Functions
Q 13 | Page 48
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