`2sin(theta-phi)=sin(theta+phi)=1`

cos2thetacos2phi+sin^2(theta-phi)-sin^2(theta+phi)=

For any two positive acute angles theta and phi if sin(theta+phi)=k sin(theta-phi) and 33(cos2phi-cos2theta)+cos2thetacos2phi=1

If sin(theta+phi)=n sin(theta-phi), then the value of (tan theta)/(tan phi) is equal to

Prove that sin^2(theta-phi)-sin^2(theta+phi)= -sin2theta sin2phi hence prove that cos2thetacos2phi-sin^2(theta+phi)+sin^2(theta-phi)=cos(2theta+2phi)

(sin(theta+phi)-2sin theta+sin(theta-phi))/(cos(theta+phi)-2cos theta+cos(theta-phi))=tan theta

Compute the following: {:[(sin(theta+phi), cos(theta+phi)),(sin(theta - phi),cos(theta-phi))] +[(sin(theta-phi), cos(theta-phi)),(sin(theta+phi), cos(theta+phi))]

prove that (cos3 theta+cos3 phi)/(2cos(theta+phi)-1)=(cos theta+cos phi)cos(theta+phi)-(sin theta+sin phi)sin(theta+phi)

(sin (theta+phi) -2 sintheta+ sin (theta-phi))/( cos(theta+phi) -2 costheta+ cos (theta-phi))= tan theta

cos2 theta cos2 phi+sin^(2)(theta-phi)-sin^(2)(theta+phi)=cos(2 theta+2 phi)

If (a-b)sin (theta+phi)=(a+b)sin(theta-phi) and a tan. (theta)/(2)-b tan.(phi)/(2)=c , then