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 θ)/(1 + tan^2 θ)) `

= `sin^(-1) (sin 2 θ)`

= 2θ

= 2 tan⁻¹ x

y = tan `phi`

तब `phi` = 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)`