From the relation `3(cos2phi-2theta)=1-cos2phicos2theta` we get `tantheta=ktanphi` Find the value of `k^4`

From the relation 3(cos2 phi-2 theta)=1-cos2 phi cos2 theta we get tan theta=k tan phi Find the value of k^(4)

Find the value of k from the following relations : 3 ( cos 2 phi - cos 2 theta) = 1- cos 2 theta cos phi , tan theta = ktan phi "where" theta and phi" are acute angles"

If 3(cos 2phi-cos2theta)=1-cos2phi cos2theta , then tan theta= k tan phi , where theta, phiin (0,(pi)/(2)) , where k =

If 3(cos 2phi-cos2theta)=1-cos2phi cos2theta , then tan theta= k tan phi , where theta, phiin (0,(pi)/(2)) , where k =

If 3(cos 2phi-cos2theta)=1-cos2phi cos2theta , then tan theta= k tan phi , where theta, phiin (0,(pi)/(2)) , where k =

Prove : (1+sin2theta-cos2theta)/(1+sin2theta+cos2theta)=tantheta

If 70sin^(2)theta+3cos^(2)theta = 4 , then tantheta

cos 2theta cos 2phi + sin^(2) (theta - phi ) - sin^(2) ( theta + phi) is equal to