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

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

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

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

Consider f(x)=tan^-1(sqrt((1+sinx)/(1-sinx)),x in(0,pi/2) . A normal to y=f(x) at x=pi/6 also passes through the point :

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

Prove that: cot^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}=pi/2-x/2 , if pi/2 < x < pi

Consider f(x) = tan^-1(sqrt((1+sinx)/((1-sinx)))), x in (0,pi/2) A normal to y = f(x) at x = pi/6 also passes through the point

Consider f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot A normal to y=f(x) at x=pi/6 also passes through the point: