Prove the following identities: 1- ^2x 1+ - ^2x 1+ =

Question

Prove the following identities: $1-\frac{\sin^2\text{x}}{1+\cot\text{x}}-\frac{\cos^2\text{x}}{1+\tan\text{x}}=\sin\text{x}\cos\text{x}$

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Answer

$\text{L.H.S}=1-\frac{\sin^2\text{x}}{1+\cot\text{x}}-\frac{\cos^2\text{x}}{1+\tan\text{x}}$ $=1-\frac{\sin^2\text{x}}{1+\frac{\cos\text{x}}{\sin\text{x}}}-\frac{\cos^2\text{x}}{1+\frac{\sin\text{x}}{\cos\text{x}}}$ $\Big(\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}},\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\Big)$ $=1-\frac{\sin^2\text{x}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}-\frac{\cos^2\text{x}}{\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}}}$ $=1-\frac{\sin^3\text{x}}{\sin\text{x}+\cos\text{x}}-\frac{\cos^3\text{x}}{\cos\text{x}+\sin\text{x}}$ $=\frac{\sin\text{x}+\cos\text{x}-(\sin^3+\cos^3\text{x})}{\sin\text{x}+\cos\text{x}}$ $=\frac{\sin\text{x}+\cos\text{x}-(\sin\text{x}+\cos\text{x})(\sin^3\text{x}+\cos^2\text{x}-\sin\text{x}\cos\text{x})}{\sin\text{x}+\cos\text{x}}$ $(\text{Using a}^3+\text{b}^3=(\text{a+b})(\text{a}^2+\text{b}^2-\text{ab}))$ $\frac{(\sin\text{x}+\cos\text{x})(1-(1-\sin\text{x}\cos\text{x}))}{\sin\text{x}+\cos\text{x}}$ $(\text{Using }\sin^2\text{x}+\cos^2\text{x}=1)$ $=\sin\text{x}\cos\text{x}$ $=\text{R.H.S}$ $\text{Proved}$

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