Question

Prove:
$\frac{cos\theta cot\theta } {1+sin\theta} = csc\theta - 1$

Answer 1

To Prove :

$\sf \dfrac{cos\theta cot\theta } {1+sin\theta} = cosec\theta - 1$

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Proof :

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LHS :

$\sf \dfrac{cos\theta cot\theta } {1+sin\theta}$

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We know that :

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

$\sf : \implies \dfrac{cos\theta \times \dfrac{cos\theta}{sin\theta}} {1+sin\theta}$

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

$\sf : \implies \dfrac{cos^{2} \theta}{sin\theta} \times \dfrac{1}{1+sin\theta}$

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By rationalisation :

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

$\sf : \implies \dfrac{cos^{2} \theta}{sin\theta} \times \dfrac{1(1-sin\theta)}{(1+sin\theta)(1-sin\theta)}$

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

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

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Now, We know that :

$\large \underline{\boxed{\bf{1 - sin^{2} \theta = cos^{2} \theta}}}$

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

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

$\sf : \implies \dfrac{1-sin\theta}{sin\theta}$

$\sf : \implies \dfrac{1}{sin\theta} - \dfrac{sin \theta}{sin\theta}$

$\sf : \implies \dfrac{1}{sin\theta} - \cancel{\dfrac{sin \theta}{sin\theta}}$

$\sf : \implies \dfrac{1}{sin\theta} - 1$

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Now, We know that :

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

$\sf : \implies cosec \theta - 1$

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RHS :

$\sf cosec\theta - 1$

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As, LHS = RHS,

Hence, proved.

Answer 2

Answer:

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