` f(x)=(1/sqrt(b-a))(((sqrt((b-a)/a))sin2x)/sqrt(1+((sqrt(b-a)/a))sinx)^2)(sqrt(a+btan^2x)` at `x=3pi/4`

f(x)=((1)/(sqrt(b-a)))(((sqrt((b-a)/(a)))sin2x)/(sqrt(1+(((sqrt(b-a))/(a)))sin x)))(sqrt(a+b tan^(2)x) at x=3(pi)/(4)(sqrt(a+b tan^(2)x)

f(x)=((1)/(sqrt(b-a)))(((sqrt((b-a)/(a)))sin2x)/(sqrt(1+(((sqrt(b-a))/(a)))sin x)))(sqrt(a+b tan^(2)x) at x=3(pi)/(4)(sqrt(a+b tan^(2)x)

If x =1/2 (sqrt(a/b)-sqrt(b/a)) then prove that (2asqrt(1+x^2))/(x+sqrt(1+x^2))= a +b

cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=(x)/(2), x in (0,(pi)/(4))

cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=(x)/(2), x in (0,(pi)/(4))

If I=int(sin x+sin^(3)x)/(cos2x)dx=P cos x+Q log|f(x)|+R then (a)P=(1)/(2),Q=-(3)/(4sqrt(2))(b)P=(1)/(4),Q=(1)/(sqrt(2)) (c) f(x)=(sqrt(2)cos x+1)/(sqrt(2)cos x-1)(d)f(x)=(sqrt(2)cos x-1)/(sqrt(2)cos x+1)

cot^(-1)[(sqrt(1-sinx)+sqrt(1+sinx))/(sqrt(1-sinx)-sqrt(1+sinx))] =a) pi-x b) 2pi-x c) x/2 d) pi-(x)/2

Prove that cot^(-1) ((sqrt(1+sin x) +sqrt(1-sin x))/(sqrt(1+sin x) -sqrt(1-sinx)))=(x)/(2), x in (0, (pi)/(4)) .

If y(x) = cot^(-1) ((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))), x in ((pi)/(2), pi) , then (dy)/(dx) at x=(5pi)/(6) is :