Question

Prove the identity:

$\rm{1-\cfrac{sin^2x}{1+cot\;x}-\dfrac{cos^2x}{1+tan\;x}=sin\;x.cos\;x}$​

Answer 1

Answer:

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Answer 2

S O L U T I O N :

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Consider LHS;

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$\bf{1-\dfrac{sin^2x}{1+cot\;x}-\dfrac{cos^2x}{1+tan\;x}}$

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$\bf{=1-\dfrac{sin^2x}{1+\dfrac{cos\;x}{sin\;x}}-\dfrac{cos^2x}{1+\dfrac{sin\;x}{cos\;x}}}$

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$\bf{=1-\dfrac{sin^3x}{sin\;x+cos\;x}-\dfrac{cos^3x}{sin\;x+cos\;x}}$

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$\bf{=\dfrac{sin\;x+cos\;x-(sin^3x-cos^3x)}{sin\;x+cos\;x}}$

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$\bf{=\dfrac{(sin\;x+cos\;x)-(sin\;x+cos\;x)(sin^2x-sin\;x\;cos\;x+cos^2x)}{sin\;x+cos\;x}}$

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$\bf{=\dfrac{(sin\;x+cos\;x)-(sin\;x+cos\;x)(1-sin\;x\;cos\;x)}{sin\;x+cos\;x}}$

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$\bf{=\dfrac{(sin\;x+cos\;x)(1-(1-sin\;x.cos\;x))}{(sin\;x+cos\;x)}}$

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$\bf{=1-(1-sin\;x.cos\;x)}$

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$\bf{=1-1+sin\;x.cos\;x}$

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$\bf{=sin\;x.cos\;x}$

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RHS;

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$\bf{=sin\;x.cos\;x}$

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∴ LHS = RHS,

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Hence, proved.