यदि `tan^(-1)((sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2))))=theta/2` है , तो `x` का मान होगा

Options

A. `sqrt(cos theta)`
B. `sqrt(sin theta)`
C. `sqrt(tan theta)`
D. `sqrt(cot theta)`

Answer

B

Explanation

`tan^(-1) ((sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))) = theta/2`
` rArr (sqrt(1+x^(2))-sqrt(1 - x^(2)))/(sqrt(1 + x^(2) )+sqrt(1 - x^(2)))= tan .theta/2`
` rArr (sqrt(1 + x^(2)))/sqrt(1 - x^(2)) = (1 + tan. theta/2)/(1 - tan .theta/2)= tan (pi/4 + theta/2)`
` rArr (1 + x^(2))/(1 - x^(2)) = tan^(2) (pi/4 + theta/2)`
` rArr 2/(2 x^(2)) = (1+tan^(2)(pi/4 + theta/2))/(tan^(2) (pi/4 + theta/2)-1)`
[योगान्तरानुपात नियम से ]
`rArr x^(2) = - cos 2 (pi/4 + theta/2)`
` [:' cos 2 theta = (1 - tan^(2) theta)/(1 + tan^(2) theta)]`
` rArr x^(2) = - cos (pi/2 + theta) rArr x = sqrt(sin theta)`