cos alpha+cos beta-a,sin alpha+sin beta-b" and "alpha-beta-2 theta" then "(cos3 theta)/(cos theta)=

If cos alpha+ cos beta= a, sin alpha+ sin beta= b and alpha- beta= 2 theta . Then (cos 3 theta)/(cos theta)=

If cos alpha+cos beta=a sin alpha+sin beta=balpha-beta=2 theta then (cos3 theta)/(cos theta)=

If cos a+cos beta=a,sin a+sin beta=b are alpha-beta=2 theta, then (cos3 theta)/(cos3 theta)=

If cos alpha + cos beta = a, sin alpha + sin beta = b and theta is the arithmetic mean between alpha and beta then sin 2 theta + cos 2 theta is equal to

If cos alpha + cos beta = 3/2 and sin alpha + sin beta = 1/2 and theta is the arithmetic mean of alpha and beta then sin 2 theta + cos 2 theta

If alpha, beta are solutions of a cos theta + b sin theta = c where a, b, c in R and a^(2) + b^(2) gt 0, cos alpha ne cos beta, sin alpha ne sin beta then prove that cos alpha cos beta = (c^(2) - b^(2))/(a^(2) + b^(2))

If alpha, beta are solutions of a cos theta + b sin theta = c where a, b, c in R and a^(2) + b^(2) gt 0, cos alpha ne cos beta, sin alpha ne sin beta then prove that cos alpha + cos beta = (2ac)/(a^(2) + b^(2))

Under rotation of axes through theta , x cosalpha + ysinalpha=P changes to Xcos beta + Y sin beta=P then . (a) cos beta = cos (alpha - theta) (b) cos alpha= cos( beta - theta) (c) sin beta = sin (alpha - theta) (d) sin alpha = sin ( beta - theta)