Question

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Answer 1

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Answer 2

Question :

$\sf \dfrac{(1+tan^{2} \theta)cot \theta}{cosec^{2} \theta} = tan\theta$

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To Prove :

• $\sf \dfrac{(1+tan^{2} \theta)cot \theta}{cosec^{2} \theta} = tan\theta$

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Proof :

$\sf \underline{\boxed{\large \bf LHS}} = \dfrac{(1+tan^{2} \theta)cot \theta}{cosec^{2} \theta}$

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We know that :

• $\large \underline{\boxed{\bf{1+tan^{2} \theta = sec^{2} \theta }}}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sec^{2} \theta cot \theta}{cosec^{2} \theta}$

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Now, we know that :

• $\large \underline{\boxed{\bf{sec \theta = \dfrac{1}{cos \theta}}}}$

• $\large \underline{\boxed{\bf{cot \theta = \dfrac{cos \theta}{sin \theta}}}}$

• $\large \underline{\boxed{\bf{sec \theta = \dfrac{1}{sin \theta}}}}$

$\sf : \implies \underline{\bf LHS} = \dfrac{\dfrac{1}{cos^{2} \theta} \times \dfrac{cos \theta}{sin \theta}}{\dfrac{1}{sin^{2} \theta}}$

$\sf : \implies \underline{\bf LHS} = \dfrac{\dfrac{1}{cos\theta \times \cancel{cos \theta}} \times \dfrac{\cancel{cos \theta}}{sin \theta}}{\dfrac{1}{sin^{2} \theta}}$

$\sf : \implies \underline{\bf LHS} = \dfrac{\dfrac{1}{cos\theta sin \theta}}{\dfrac{1}{sin^{2} \theta}}$

$\sf : \implies \underline{\bf LHS} = \dfrac{1}{cos\theta sin \theta} \times sin^{2} \theta$

$\sf : \implies \underline{\bf LHS} = \dfrac{1}{cos\theta \cancel{sin \theta}} \times \cancel{sin \theta} \times sin \theta$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin \theta}{cos \theta}$

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Now, we know that :

• $\large \underline{\boxed{\bf{\dfrac{sin \theta}{cos \theta} = tan \theta}}}$

$\sf : \implies \underline{ \bf LHS} = tan \theta$

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$\sf \underline{\boxed{\large \bf RHS}} = tan \theta$

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$\large \underline{\boxed{\sf As, \: LHS = RHS,}}$

$\large \underline{\boxed{\sf Hence, \: proved.}}$