The area enclosed by the curve `y=sinx+cosxa n dy=|cosx-sinx|` over the interval `[0,pi/2]` is `(a)4(sqrt(2)-2)` (b) `2sqrt(2)` (`sqrt(2)` -1) `(c)2(sqrt(2)` +1) (d) `2sqrt(2)(sqrt(2)+1)`

The value of the definite integral int_0^(pi/2)sqrt(tanx)dx is (a) sqrt(2)pi (b) pi/(sqrt(2)) (c) 2sqrt(2)pi (d) pi/(2sqrt(2))

sinx + cosx = sqrt(2) cos2x

Find the angles of the triangle whose sides are (sqrt(3)+1)/(2sqrt(2)), (sqrt(3)-1)/(2sqrt(2)) and sqrt(3)/2 .

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)

sinx + cosx = 1/sqrt(2)

Prove that tan(pi/(16))=sqrt(4+2sqrt(2))-(sqrt(2)+1)

int_(0)^((pi)/(2))(xdx)/(sinx+cosx)=(pi)/(2sqrt(2))log(sqrt(2)+1)

The value of x in (0,pi/2) satisfying (sqrt(3)-1)/(sinx)+(sqrt(3)+1)/(cosx)=4sqrt(2) is / are

If cosx=(2cosy-1)/(2-cosy), where x , y in (0,pi) then tan(x/2)xxcot(y/2) is equal to (a) sqrt(2) (b) sqrt(3) (c) 1/(sqrt(2)) (d) 1/(sqrt(3))

A(1/(sqrt(2)),1/(sqrt(2))) is a point on the circle x^2+y^2=1 and B is another point on the circle such that arc length A B=pi/2 units. Then, the coordinates of B can be (a) (1/(sqrt(2)),-1/sqrt(2)) (b) (-1/(sqrt(2)),1/sqrt(2)) (c) (-1/(sqrt(2)),-1/(sqrt(2))) (d) none of these