General solution of `(i) sin^2theta = sin^2alpha; (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 (i)sin^(2)theta=sin^(2)alpha;(ii)cos^(2)theta=cos^(2)alpha(iii)tan^(2)theta=tan^(2)alpha

Show that the three equations sin^(2)theta = sin^(2)alpha, cos^(2)theta = cos^(2)alpha and tan^(2)theta = tan^(2)alpha are all identical and the solution of each equation is npi +-alpha .

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

Show that the equations, sin^2theta=sin^2alpha,cos^2theta=cos^2alpha and tan^2theta=tan^2alpha are same and the solution of each of them are npi+-alpha

If cos^2theta-sin^2 theta = tan^2 alpha , then show that cos^2 alpha -sin^2 alpha = tan^2 theta .

If sin ( theta + alpha ) = cos ( theta + alpha ) , then tan theta =

If sin theta = nsin(theta +2alpha) then tan(theta + alpha) =

Evaluate int (cos 2 theta - cos 2 alpha )/(sin alpha - sin theta ) d theta

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