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)

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

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

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

Taking theta=30^(@) , verify each of the following (i) sin 2theta=2sin theta cos theta (ii) cos theta=2cos^(2)theta-1=1-2sin^(2)theta (iii) tan2theta=(2tantheta)/(1-tan^(2)theta)

tan theta = (sin alpha-cos alpha) / (sin alpha + cos alpha)

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 ____________.

If sin theta =3 sin ( theta + 2 alpha), then the value of tan (theta + alpha) + 2 tan alpha is