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: $$ \frac{(\frac{\cos^2\alpha}{a^2}+\frac{\sin^2\alpha}{b^2})-(\frac{m²\sin^2\alpha}{a^2}+\frac{m²\cos^2\alpha}{b^2})}{m(2\sin\alpha\cos\alpha)(\frac{1}{a²}-\frac{1}{b²})}=\frac{\cot\alpha(m²a²-b²)-\tan\alpha(a²-m²b²)}{2m(a²-b²)} $$
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$$RHS=\frac{\cos^2\alpha(m^2a^2-b^2)-\sin^2\alpha(a^2-m^2b^2)}{2\sin\alpha\cos\alpha. m(a^2-b^2)}$$
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
answered May 5 '17 at 1:29
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$\begingroup$ Thak you very much, dividing by a^2b^2 was the answer. $\endgroup$ – P Macmutton May 5 '17 at 11:49
either I am missing some trigonometric identityMaybe $\tan \alpha = \sin \alpha / \cos \alpha$ and $\cot \alpha = \cos \alpha / \sin \alpha$. For example $\cos^2 \alpha / (\sin \alpha \cos \alpha) = \cos \alpha / \sin \alpha = \cot \alpha$. $\endgroup$ – dxiv May 5 '17 at 1:26