How to establish the identity

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How do I establish the identity in this problem? Struggling with this one at the moment. $$\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\tan\theta}{1-\tan^2\theta}$$

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4 Answers

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$\bf hint: $ divide the top and bottom of $$\frac{\sin t\cos t}{\cos^2 t- \sin^2 t } $$ by $\cos^2 t.$ now use the fact $\tan t = \frac{\sin t}{\cos t}.$

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We have $$\frac{\tan \theta}{1-\tan^2 \theta}=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{\sin\theta(\cos^2\theta)}{\cos\theta(\cos^2\theta-\sin^2\theta)}=\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}$$

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Put $\tan = \frac{\sin}{\cos}$ and simplify.

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$$\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\tan\theta}{1-\tan^2\theta}$$ $$\implies \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{2\tan\theta}{1-\tan^2\theta}$$ Now using $\sin 2\theta=2\sin\theta\cos\theta$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$ $$\frac{\sin 2\theta}{\cos 2\theta}=\tan 2\theta$$ Now using $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ we get $$\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$$ Which is what we wanted to prove.

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