If `2sinx=1`,`(pi/2)``<``x``<``pi` and `sqrt2` cosy=1, `((3pi)/2)``<``y``<``2pi` then find the value of `(frac(tanx+tany)(cosx-cosy))`

If sinx=(2t)/(1+t^2) , tany=(2t)/(1-t^2) , find (dy)/(dx) .

Find the equations of tangents and normals to the curve at the point on it: 2xy + pi sin y = 2 pi at (1, (pi)/2)

If f(x) is continuous at x = (pi)/2 , where f(x) = (sinx)^(1/(pi-2x)), "for" x != (pi)/2 , then f((pi)/(2)) =

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:

Prove that (i) " 2sin " (5pi)/(12) " sin " (pi)/(12)=(1)/(2) (ii) " 2 cos " (5pi)/(12) " cos " .(pi)/(12)=(1)/(2) (iii) " 2 sin ".(5pi)/(12) " cos " (pi)/(2) = ((2+sqrt(3))/(2))

The value of int_(-pi//2)^(pi//2)(1)/(e^(sinx)+1) dx is equal to

The value of int _(0)^(pi//2) (2^(sinx))/(2^(sinx)+2^(cosx))dx is

If f(x)=|(sin x+sin 2x+sin3x,sin2x,sin3x),(3+4sinx,3,4sinx),(1+sinx,sinx,1)|, then the value of int_0^(pi/2) f(x) dx is