If `sqrt((1-sinx)/(1+sinx))= f(pi/4-x/2)` then `f=`

sqrt((1+sinx)/(1-sinx))=tan(pi/4+x/2)

inttan^(- 1)sqrt((1-sinx)/(1+sinx))dx ,-pi/2

If f(x) = tan^(-1)(sqrt((1+sinx)/(1-sinx))), 0 lt x lt pi/2 , then f'(pi/6) is

Differentiate tan^(-1){sqrt((1+sinx)/(1-sinx))}, -pi/2

If f(x)=(sqrt(1+sinx)-sqrt(1-sinx))/(x) , then we value of f at x = 0, so that f is continuous everywhere, is

Evaluate: inttan^(-1){sqrt(((1-sinx)/(1+sinx)))} dx , -pi//2

If f(x) is continuous at x=pi/2 , where f(x)=(cos x )/(sqrt(1-sinx)) , for x!= pi/2 , then f(pi/2)=

The value of tan^(-1){(sqrt(1-sinx)+sqrt(1+sinx))/(sqrt(1-sinx)-sqrt(1+sinx))} is : ((pi)/(2) lt x lt pi)