tan^(2)((pi x^(2))/(2))

int ["tan"(x)/(2)+"tan" ((pi)/(2)-(x)/(2))]dx=

Compute the area of the region bounded by the curves y=tan x and y=tan^(2)x(-(pi)/(2)

(sec2x-tan2x) equals a tan(x-(pi)/(4))b)tan((pi)/(4)-x)c cot(x-(pi)/(4))d tan^(2)(x+(pi)/(4))

int _ (-2)^(1) (tan^(-1) ((x)/(x^(2) +1))+tan^(-1) ((x^(2) +1)/( x))) dx is (i) (pi)/(2) (ii)-(pi)/(2) (iii) pi (iv) -pi

If 0 lt x lt (pi)/(4), then sec 2x - tan 2x= ................ A) tan (x- (pi)/(4)) B) tan ((pi)/(4) -x) C) tan (x+(pi)/(4)) D) tan ^(2) (x+ (pi)/(4))

tan^(2)((pi)/(16))+tan^(2)((2 pi)/(16))+tan^(2)((3 pi)/(16))... ..+tan^(2)((7 pi)/(16)/)=35

if tan ((pi)/(4)+(y)/(2))=tan^(3)((pi)/(4)+(x)/(2)) then (3+sin^(2)x)/(1+3sin^(2)x)=

If tan((pi)/(4)+(y)/(2))=tan^(3)((pi)/(4)+(x)/(2)) then (3+sin^(2)x)/(1+3sin^(2)x)=

tan((pi)/(4)+(x)/(2))+tan((pi)/(4)-(x)/(2))=2 then the general solution of x=