The expression `(1+cosx+cos2x)/(sinx+sin2x)` equals ; `x in (0,pi/2)`

sinx + cosx = (cos2x)/(1-sin2x)

cosx + sinx = cos2x + sin2x

If Delta (x)=|{:(1,cos x,1-cos x),(1+sin x,cos x,1+sinx-cosx),(sinx,sinx,1):}| then int_(0)^(pi//2)Delta(x) dx is equal to

If Delta (x)=|{:(1,cos x,1-cos x),(1+sin x,cos x,1+sinx-cosx),(sinx,sinx,1):}| then int_(0)^(pi//2)Delta(x) dx is equal to

If Delta (x)=|{:(1,cos x,1-cos x),(1+sin x,cos x,1+sinx-cosx),(sinx,sinx,1):}| then int_(0)^(pi//2)Delta(x) dx is equal to

Solve: 2(sinx -cos2x)-sin2x(1+2sinx) + 2cosx =0

Assertion : If 0ltxlt(pi)/2 then sin^(-1)(cosx)+cos^(-1)(sinx)=pi-2x Reason : cos^(-1)x=(pi)/2-sin^(-1)x,AA x in [-1,1]

Solve sin2x-cos2x=sinx-cosx

Let f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|, the value of int_0^(pi//2){f(x) + f'(x)} dx, is