Find `sin2x,cos2x,tan2x`if `secx=-13/5` , `pi/2``<``x``<``pi`

Let B=2sin^2x-cos2x , then

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

2 int frac{cos^2 x - sin^2 x}{cos^2 x + sin^2 x} dx=............ A) sin 2x + c B) cos 2x + c C) tan 2x + c D) 2sin 2x + c

Principal solutions of the equation sin 2x + cos 2x = 0 , where pi < x < 2pi

If the general solution of sin5x=cos2x is of the form a_n.pi/2 for n=0,+-1,+-2,.... then a_n=

Find the equations of tangents and normals to the curve at the point on it: x sin 2y = y cos 2x at ((pi)/4, (pi)/2)

If int (sin2x-cos2x)dx=1/sqrt2sin(2x-a)+b then

If 2sin^2[(pi/2)cos^2x]=1-cos(pisin2x) , x!=(2n+1)pi/2,nin1 , then cos 2x is equal to

If x in [0. (pi)/(2)], y in [0, (pi)/(2)] and sin x + cos y =2, then the value of x+y is equal to........ A) 2pi B) pi C) pi/4 D) pi/2