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Prove that $\tan^2\theta -\sin^2 \theta =\tan^2\theta.\sin^2\theta $
in Mathematics by Wooden (4,853 points)
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To prove that $\tan^2\theta -\sin^2 \theta =\tan^2\theta.\sin^2\theta $

Lets take the left hand side

since $\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$ we have

\[\tan^2\theta -\sin^2 \theta\]

\[=\dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta\]

combine into one fraction

\[=\dfrac{\sin^2\theta - \sin^2\theta \cdot \cos^2\theta}{\cos^2\theta}\]

In the numerator, take out the common factor $\sin^2\theta$

\[\dfrac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}\]

but $1-\cos^2\theta$ is the same as $\sin^2\theta$, so we can substitute

\[\dfrac{\sin^\theta \cdot \sin^2\theta}{\cos^2\theta}\]

since $\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$ we have

\[\tan^2\theta \cdot \sin^2\theta\]

which is the right hand side :)
by Wooden (4,853 points)

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