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

Options

A. ` sqrt((sqrt5+1)/2)`
B. `sqrt((1-sqrt5)/2)`
C. `sqrt((sqrt5-1)/2)`
D. `sqrt((1+sqrt5)/3)`

Answer

C

Explanation

दिए है , `sin^(-1) x + tan^(-1) x = pi/2`
हम जानते है कि ,
` sin^(-1) x = tan^(-1)(x/(sqrt(1 - x^(2))))`
`:. tan^(-1)(x/(sqrt(1-x^(2)))) + tan^(-1) x = pi/2` परन्तु `tan^(-1) x + tan^(-1) y = tan^(-1)((x+y)/(1-xy))`
` :. tan^(-1){(x/sqrt(1-x^(2))+x)/(1-x/(sqrt(1-x^(2)))*x)}= pi/2`
` rArr (x+xsqrt(1-x^(2)))/(sqrt(1-x^(2))-x^(2))=infty = 1/0`
अतः `sqrt(1-x^(2))-x^(2) = 0`
` rArr (sqrt(1-x^(2)))^(2) = (x^(2))^(2)`
तब ,`x^(2) = (-1pmsqrt(1+4))/(2(1))rArr x^(2) = (-1pmsqrt5)/2`
अतः ` x = sqrt((sqrt5-1)/2)`