Question

question is in photo answer correctly​

Answer 1

Question :-

$\large \bullet \sf \blue{ \frac{tan \theta - cot \theta}{sin \theta.cos \theta} = sec {}^{2} \theta - cosec {}^{2} \theta}$

Solution :-

LHS :-

$\implies\: \large \sf \orange{ \frac{tan \theta - cot \theta}{sin \theta.cos \theta}}$

$\implies \: \large \sf \large \: \frac{ \frac{sin \theta}{cos \theta} - \frac{cos \theta}{sin \theta} }{sin \theta.cos \theta}$

$\implies \large \: \sf \: \frac{sin {}^{2} \theta - cos {}^{2} \theta }{(sin \theta.cos \theta)(cos \theta .sin \theta)}$

$\implies \large \sf \: \frac{sin {}^{2} \theta - cos {}^{2} \theta }{sin {}^{2} \theta.cos {}^{2} \theta }$

$\implies \large \sf \: \frac{ sin {}^{2} \theta }{sin {}^{2} \theta.cos {}^{2} \theta } - \frac{cos {}^{2} \theta }{sin {}^{2} \theta.cos {}^{2} \theta }$

$\implies \sf \large \: \frac{1}{cos {}^{2} \theta } - \frac{1}{sin {}^{2} \theta }$

$\implies \sf \large \red{sec {}^{2} \theta - cosec {}^{2} \theta}$

RHS.

Hence,

$\large \bullet \sf \green{ \frac{tan \theta - cot \theta}{sin \theta.cos \theta} = sec {}^{2} \theta - cosec {}^{2} \theta}$

Answer 2

Identity :-

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$\sf{\bold{\dfrac{ tan\theta - cot\theta}{sin\theta . cos\theta } = sec^2\theta - cos^2\theta}}$

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LHS:-

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: $\sf\implies \: {\bold {\dfrac{ tan\theta - cot\theta } { sin\theta . cos\theta}}}$

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• $\sf{\red{\boxed{\bold{tan\theta = \dfrac{sin\theta}{cos\theta}}}}}$

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• $\sf{\red{\boxed{\bold{cot\theta = \dfrac{cos\theta}{sin\theta}}}}}$

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: $\sf\implies \: {\bold {\dfrac{ \dfrac{sin\theta}{cos\theta} - \dfrac{cos\theta}{sin\theta} } { sin\theta . cos\theta}}}$

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: $\sf\implies \: {\bold {\dfrac{ \dfrac{sin^2\theta - cos^2\theta}{sin\theta. cos\theta} } { sin\theta . cos\theta}}}$

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: $\sf\implies \: {\bold {{ \dfrac{sin^2\theta - cos^2\theta}{sin\theta. cos\theta} } \div { sin\theta . cos\theta}}}$

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: $\sf\implies \: {\bold {{ \dfrac{sin^2\theta - cos^2\theta}{sin\theta. cos\theta} } \times \dfrac{1} { sin\theta . cos\theta}}}$

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: $\sf\implies \: {\bold {\dfrac{ sin^2\theta - cos^2\theta } { sin^2\theta . cos^2\theta}}}$

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: $\sf\implies \: {\bold {\dfrac{\cancel{sin^2\theta}}{\cancel{sin^2\theta } . cos^2\theta} - \dfrac{\cancel{cos^2\theta}}{ sin^2\theta . \cancel{cos^2\theta }} }}$

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: $\sf\implies \: {\bold{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}}$

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• $\sf{\red{\boxed{\bold{sec^2\theta= \dfrac{1}{cos^2}}}}}$

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• $\sf{\red{\boxed{\bold{cosec^2\theta= \dfrac{1}{sin^2}}}}}$

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: $\sf\implies \: {\bold{ sec^2\theta - cosec^2\theta }}$

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RHS :-

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: $\sf\implies \: {\bold{ sec^2\theta - cosec^2\theta }}$

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Compare:-

$\sf \: {\bold{ sec^2\theta - cosec^2\theta }}$ = $\sf \: {\bold{ sec^2\theta - cosec^2\theta }}$

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......Hence proved

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