Question

निम्नलिखित प्रश्न का मान ज्ञात कीजिए:

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

Answer 1

x = tan θ. , तब 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`

y = tan Φ. तब, 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)`