can someone help me prove this identity? (tan^2x)-(sin^2x)=(sin^2x)(tan^2x)
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Originally Posted by wisezeta can someone help me prove this identity? (tan^2x)-(sin^2x)=(sin^2x)(tan^2x) Hi Start proving from the right hand side $\displaystyle \tan^2x-\sin^2x=\frac{\sin^2x}{\cos^2x}-\sin^2x$ $\displaystyle =\frac{\sin^2x(1-\cos^2x)}{\cos^2x}$ $\displaystyle =\frac{\sin^2x}{\cos^2x}(\sin^2x)$ $\displaystyle =\tan^2x\sin^2x $ Proved .
(tan^2x)-(sin^2x)=(sin^2x)(tan^2x) an identity treated as an equation, the graph is shown below
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