If `sinx+cosx=sqrt(y+1/y)` for `x in [0,pi]` , then `x=pi/4` (b) `y=0` `y=1` (d) `x=(3pi)/4`

If sinx+cosx=sqrt(y+1/y) for x in [0,pi] , then (a) x=pi/4 (b) y=0 (c) y=1 (d) x=(3pi)/4

If sinx+cosx=sqrt(y+1/y) for x in [0,pi] , then (a) x=pi/4 (b) y=0 (c) y=1 (d) x=(3pi)/4

If sin x+cos x=sqrt(y+(1)/(y)) for x in[0,pi] then x=(pi)/(4) (b) y=0y=1 (d) x=(3 pi)/(4)

If sinx+cosx=sqrt(y+1/y), y gt 0, x in[0,pi] , then find the least value of x satisfying the given conditions.

If sinx+cosx=sqrt(y+1/y), y gt 0, x in[0,pi] , then find the least value of x satisfying the given conditions.

If x^2-4x+5-siny=0, yepsilon [0,2pi] then (A) x=1,y=0 (B) x=1,y=pi/2 (C) x=2,y=0 (D) x=2,y=pi/2

If xa n dy are positive acute angles such that (x+y) and (x-y) satisfy the equation tan^2theta-4tantheta+1=0, then x=pi/6 (b) y=pi/4 (c) y=pi/6 (d) y=pi/4

If x^(2)-4x+5-siny=0, y in [0,2pi] , then- A) x=1, y=0 , B) x=1, y=pi/2 , C) x=2, y=0 , x=2, y=pi/2

The area of the region between the curves y= sqrt((1+sinx)/(cosx))and y = sqrt((1-sinx)/(cosx)) bounded by the lines x = 0 and x = pi/4 is

If f(x)=sqrt(1-sin2x) , then f^(prime)(x) is equal to (a) -(cosx+sinx) ,for x in (pi/4,pi/2) (b) cosx+sinx ,for x in (0,pi/4) (c) -(cosx+sinx) ,for x in (0,pi/4) (d) cosx-sinx ,for x in (pi/4,pi/2)