It might be useful to solve for each side separately until they converge while taking note of how each part looks! For example:
a) RHS is easily factorisable and LHS can be expressed in terms of cos2, so after taking those two out of the picture it would be simpler to find the connection between the complex RHS and eventually getting it to be equal to sin2)!
b) RHS should motivate you to obtain some form of individual terms that separate into the terms from LHS, which is exactly what happens, since you can see how the sin/cos is equivalent to the tan component, which leaves just the pythagorean identity
c) LHS is relatively simple, so it would be unproductive to start there, instead, finding a simplification for RHS by noticing that they fulfill the form (1+a)(1-a) = 1-a2 helps you cancel out many terms that end up simplifying to sin/cos!
d) Similar to (b), the cos/sincos should simplify to the eventual cosec, which shows that the remaining bit should simplify to tan+cot!