One side of a square makes an angle `alpha` with x axis and one vertex of the square is at origin. Prote that the equations of its diagonals are `x(sin alpha+ cos alpha) =y (cosalpha-sinalpha)` or `x(cos alpha-sin alpha) + y (sin alpha + cos alpha) = a`, where a is the length of the side of the square.

tan theta = (sin alpha-cos alpha) / (sin alpha + cos alpha)

(sqrt2 - sin alpha - cos alpha )/( sin alpha - cos alpha)=

A square lies above the X - axis and has one vertex at the origin . The side passing through the origin makes an angle alpha(o lt alpha lt pi//4) with the positive direction of the X - axis . Prove that the equation of its diagonals are , y (cos alpha - sin alpha ) = x (sin alpha + cos alpha ) , and y(sin alpha + cos alpha ) + x ( cos alpha - sin alpha ) = a where , is the length of each side of the square

If A_(alpha), = [{:(cos alpha , sin alpha),(-sin alpha , cos alpha):}] , then

((cos alpha-cos3 alpha)(sin8 alpha+sin2 alpha))/((sin5 alpha-sin alpha)(cos4 alpha-cos6 alpha))=

If tan theta=(sin alpha- cos alpha)/(sin alpha+cos alpha) , then

If x=a sin alpha+b cos alpha and y=a cos alpha-b sin alpha then remove alpha

If (2sin alpha) / ({1 + cos alpha + sin alpha}) = y, then ({1-cos alpha + sin alpha}) / (1 + sin alpha) =