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Tan^2xsin^2x=tan^2x-sin^2x
Prove identity

Sagot :

Answer:

See below for proof

Step-by-step explanation:

[tex]\tan^2x\sin^2x\\\\(\sec^2x-1)(\sin^2x)\\\\(\frac{1}{\cos^2x}-\frac{\sin^2x}{\sin^2x})(\sin^2x)\\\\(\frac{\sin^2x-\sin^2x\cos^2x}{\sin^2x\cos^2x})(\sin^2x)\\\\\frac{\sin^2x-\sin^2x\cos^2x}{\cos^2x}\\\\\frac{\sin^2x}{\cos^2x}-\sin^2x\\\\\tan^2x-\sin^2x[/tex]

Hence, the identity is proven

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