# Help simplifying a complex trigonometric equation

I have problems to understand how can I go from the left-hand side to the right-hand side of the following equation. The right-hand side is supposed to result from simplification of the left-hand side, but either I am missing some trigonometric identity, or an algebraic step, in any case help is welcome: $$\frac{(\frac{\cos^2\alpha}{a^2}+\frac{\sin^2\alpha}{b^2})-(\frac{m²\sin^2\alpha}{a^2}+\frac{m²\cos^2\alpha}{b^2})}{m(2\sin\alpha\cos\alpha)(\frac{1}{a²}-\frac{1}{b²})}=\frac{\cot\alpha(m²a²-b²)-\tan\alpha(a²-m²b²)}{2m(a²-b²)}$$

• either I am missing some trigonometric identity Maybe $\tan \alpha = \sin \alpha / \cos \alpha$ and $\cot \alpha = \cos \alpha / \sin \alpha$. For example $\cos^2 \alpha / (\sin \alpha \cos \alpha) = \cos \alpha / \sin \alpha = \cot \alpha$. – dxiv May 5 '17 at 1:26
• thanks, the problem was lack of ability for algebra more than not knowing those identities! – P Macmutton May 5 '17 at 11:48

$$RHS=\frac{\cos^2\alpha(m^2a^2-b^2)-\sin^2\alpha(a^2-m^2b^2)}{2\sin\alpha\cos\alpha. m(a^2-b^2)}$$
Using:$\tan\alpha=\sin\alpha/\cos\alpha$ and $\cot\alpha=(\tan\alpha)^{-1}$ Rest, divide by $a^2b^2$ in numerator and denominator to get your answer upon simplification