Prove that `(i) sin^2theta = sin^2alpha; (ii) cos^2 theta = cos^2 alpha (iii) tan^2 theta = tan^2 alpha`

Prove that (i)sin^(2)theta=sin^(2)alpha;(ii)cos^(2)theta=cos^(2)alpha(iii)tan^(2)theta=tan^(2)alpha

General term ( sin^2 theta = sin^2 alpha , cos^2 theta = cos^2 alpha, tan^2 theta= tan^2 alpha)

Prove that (sin theta - 2 sin^(3) theta)/(2 cos^(3) theta - cos theta) = tan theta

Solution ( Principle & General Solution ) || General form ( tan theta = tan alpha, cos theta = cos alpha , sin theta = sin alpha)

Prove that cos theta =(cos alpha- cos beta)/(1 -cos alpha*cos beta) ⇔ tan theta/2 = pm tan alpha/2 *cot beta/2 .

If : cos 2 theta = sin alpha , then : theta is given by

Prove that: (sin theta - 2 sin^(3) theta) = (2cos^(3) theta - cos theta) tan theta .

If cos^(2)theta-sin^(2)theta=tan^(2)alpha, "then" : sqrt2 * cos theta * cos alpha=.... A) sin theta * sin alpha B)1 C) tan^(2)theta D)none of these

If ( sin ^(3) theta)/( sin ( 2theta+ alpha )) = ( cos ^(3) theta)/( cos ( 2 theta + alpha)) and tan 2 theta = lamda tan ( 3theta + alpha) then the value of lamda is ____________.

(i) tan3 theta tan7 theta + 1 = 0 (ii) sin ^ (2) theta = sin ^ (2) alpha