If $y=3\tan^3{\frac{x}{2}}$ and $y'=a\sin{bx} \cos{cx} $

Question

I want to solve the following problem. If $y=3\tan^3{\frac{x}{2}}$ and $y'=a\sin{bx} \cos{cx} $, then $\frac{a} {bc} =$

What I have done so far, $y'=\frac{9}{2} \tan^2{ \frac{x}{2}} \sec^2{\frac{x}{2} }$

But I don't know, how change that to $sin$ and $cos$ term

Answer 1

I have get $$y'=9\, \left( \tan \left( x/2 \right) \right) ^{2} \left( 1/2+1/2\, \left( \tan \left( x/2 \right) \right) ^{2} \right) $$ this simplifies to $$y'={\frac {9-9\,\cos \left( x \right) }{ \left( \cos \left( x \right) +1 \right) ^{2}}} $$ multiplying denominator and numerator by $$1+\cos(x)$$ we get $$\frac{9\sin^2(x)}{(1+\cos(x))^3}$$