Prove that : `cos^2alpha+cos^2(alpha+beta)-2cosalphacosbetacos(alpha+beta)=sin^2beta`

The value of the determinant |{:(1,sin(alpha-beta)theta,cos (alpha-beta)theta),(a, sinalphatheta,cos alphatheta),(a^(2),sin(alpha-beta)theta,cos(alpha-beta)theta):}| is independent of

cos^(2) alpha +cos^(2) beta +cos^(2) gamma is equal to

Let y = f(x), f : R ->R be an odd differentiable function such that f'''(x)>0 and g(alpha,beta)=sin^8alpha+cos^8beta+2-4sin^2alpha cos^2 beta If f''(g(alpha, beta))=0 then sin^2alpha+sin^2beta is equal to

If cosalpha+cosbeta=0=sinalpha+sinbeta, then prove that cos2alpha+cos2beta=-2cos(alpha+beta) .

If cot (alpha + beta )=0 , then sin(alpha+2beta ) =

Prove that 2tanbeta+cotbeta=tanalphaimplies2tan(alpha-beta)=cotbeta .

If sin(theta+alpha)=aandsin(theta+beta)=b , then prove that cos2(alpha-beta)-4abcos(alpha-beta)=1-2a^(2)-2b^(2) .

If alphaandbeta are not the multiple of (pi)/(2)and [{:(cos^(2)alpha,cosalphasinalpha),(cosalphasinalpha,sin^(2)alpha):}]xx[{:(cos^(2)beta,sinbetacosbeta),(sinbetacosbeta,sin^(2)beta):}]=[{:(0,0),(0,0):}] then alpha-beta is ......

If alpha,beta are roots of x^2-px+q=0 then find the quadratic equation whose roots are ((alpha^2-beta^2)(alpha^3-beta^3)) and alpha^2beta^3+alpha^3beta^2