Question
Verify \tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x =\dfrac{2\sin x}{\cos x+\cos 2x}
Verify \tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x =\dfrac{2\sin x}{\cos x+\cos 2x}
Using the sum to product identity for tangent function,
\tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x
=\dfrac{\sin(\dfrac{3}{2}x-\dfrac{1}{2}x )}{\cos\dfrac{3}{2}x\cos \dfrac{1}{2}x }
Using product to sum identity
\cos \alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]
\tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x
=\dfrac{2\sin x}{\cos x \cos 2x}