If `veca=(3,1) and vecb=(1,2)` represent the sides of a parallelogram then the angle `theta` between the diagonals of the paralelogram is given by (A) `theta=cos^-1(1/sqrt(5))` (B) `theta=cos^-1(2/sqrt(5))` (C) `theta=cos^-1 (1/(2sqrt(5)))` (D) `theta = pi/2`

cos((sqrt(2)+1)/(2))theta cos((sqrt(2)-1)/(2))theta=1

cos2 theta-(sqrt(2)+1)(cos theta-(1)/(sqrt(2)))=0

If cos theta = (1)/(sqrt2) " then " theta

If sin theta=(1)/(sqrt(2)) and cos theta=(1)/(sqrt(2)), then cot theta=?

tan theta=-(1)/(sqrt(3)),sin theta=(1)/(2) and cos theta=-(sqrt(3))/(2)

If cos 2 theta =(sqrt(2)+1)( cos theta -(1)/(sqrt(2))) , then the value of theta is

Solve the following cos2 theta=(sqrt(2)+1)(cos theta-(1)/(sqrt(2)))

The most general value of theta satisfying the equations sin theta=(1)/(sqrt(2)) and cos theta=-(1)/(sqrt(2)) is

If the tangent drawn at point (t^(2),2t) on the parabola y^(2)=4x is the same as the normal drawn at point (sqrt(5)cos theta,2sin theta) on the ellipse 4x^(2)+5y^(2)=20 ,then theta=cos^(-1)(-(1)/(sqrt(5)))( b) theta=cos^(-1)((1)/(sqrt(5)))t=-(2)/(sqrt(5))(d)t=-(1)/(sqrt(5))