Question

Sin²x tan²x = tan²x - sin²x

Answer 1

$\displaystyle \sin^2x\tan^2x=\tan^2x-\sin^2x\\\sin^2x\cdot\frac{\sin^2x}{\cos^2x}=\tan^2x-\sin^2x\\\frac{\sin^2x\cdot\sin^2x}{\cos^2x}=\tan^2x-\sin^2x\\\frac{\sin^2x\cdot(1-\cos^2x)}{\cos^2x}=\tan^2x-\sin^2x\\\frac{\sin^2x-\cos^2x\sin^2x}{\cos^2x}=\tan^2x-\sin^2x\\\frac{\sin^2x}{\cos^2x}-\sin^2x=\tan^2x-\sin^2x\\\boxed{\boxed{\tan^2x-\sin^2x=\tan^2x-\sin^2x}}$

Answer 2

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