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I have to prove that: $$\tan^2\theta \sin^2\theta = \tan^2\theta - \sin^2 \theta$$ Here is what I have tried $$\tan^2\theta \sin^2\theta$$ $$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$ $$=\frac{\sin^4\theta}{\cos^2\theta}$$ Not much of an attempt, but now I am stuck. What should I do next? Thanks in advance for your answers ;)

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$\sin^4 \theta = \sin^2 \theta \sin^2 \theta = (1 - \cos^2\theta)\sin^2\theta = \sin^2\theta - \sin^2\theta \cos^2\theta$. –  Tunococ Mar 14 at 0:38

3 Answers 3

up vote 2 down vote accepted

$$\tan^2\theta \sin^2\theta$$ $$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$ $$=\frac{\sin^4\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta\sin^2\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}$$ $$=\frac{\sin^2\theta-\sin^2\theta\cos^2\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta}{\cos^2\theta}-\frac{\sin^2\theta\cos^2\theta}{\cos^2\theta}$$ $$=\tan^2\theta-\sin^2\theta$$ $$\displaystyle \boxed{\therefore \tan^2\theta \sin^2\theta=\tan^2\theta - \sin^2\theta}$$

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$$\tan^2\theta \sin^2\theta+\sin^2 \theta=\sin^2\theta(\tan^2\theta +1) = \sin^2\theta\cdot\sec^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$


Alternatively, $$\frac1{ \sin^2\theta}-\frac1{\tan^2\theta}=\csc^2\theta-\cot^2\theta=1$$

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@user135339, how about this? –  lab bhattacharjee Mar 14 at 3:18

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

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