Proved the following identities:
2. cos4a. tan2a-sin4a=(2tan a)/(tan^2 a -1)
Wouldn't it be better for you to proe them? Do you know identities for trig functions of multiple or fractional arguments?
In the second, we can replace by to get
so that , and letting that becomes . Similarly, and .
Those, together with the fact that , are what you need.
The second one is quite simple
LHS = cos 4a tan 2a - sin 2a = ( cos 4a sin 2a ) / cos 2a - sin 2a = ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a
RHS = ( 2 tan a )/ ( tan^2 a - 1 ) = - ( 2 tan a )/ ( 1 - tan^2 a ) = - tan 2a
hence LHS = RHS