Question

# Quality Question
@Trigonometry

Prove That

$\bf \dfrac{cos \theta}{1 - tan \theta} + \frac{sin \theta}{1 - co t \theta} = cos \theta + sin \theta$
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Answer 1

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To Prove :-

$\mathtt{ \dfrac{ cos \theta}{1 - tan \theta} + \dfrac{ sin \theta}{1 - cot \theta} = cos \theta + sin \theta}$

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▪ Formulae Used :-

$\maltese \: \: \bf tan\theta=\dfrac{sin\theta}{cos\theta} \\ \\ \maltese \: \: \:\bf cot\theta=\dfrac{cos\theta}{sin\theta} \\ \\\maltese \: \: \:\bf a^2-b^2=(a+b)(a-b)$

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▪Proof :-

$\large\mathcal{LHS} :\\ \\ \mathtt{ \dfrac{ cos \theta}{1 - tan \theta} + \dfrac{ sin \theta}{1 - cot \theta} } \\ \\ = \: \mathtt{ \dfrac{cos \theta}{1 - \dfrac{sin\theta}{cos\theta } } + \dfrac{sin\theta}{1 - \dfrac{cos\theta}{sin\theta} } } \\ \\ \: =\mathtt{ \frac{cos {}^{2} \theta}{cos\theta - sin\theta} + \frac{ {sin}^{2} \theta}{sin\theta - cos\theta} } \\ \\ \: = \mathtt{ \frac{ {cos}^{2}\theta - {sin}^{2} \theta }{cos\theta - sin\theta} } \\ \\ \: = \mathtt{ \frac{(cos\theta - sin\theta)(cos\theta + sin\theta)}{cos\theta - sin\theta} } \\ \\ \: = \mathtt{cos \theta +sin\theta }=\large\mathcal{RHS}$

$\huge\color{magenta} \red{\bigstar}\:\: \mathfrak{\text{H}ence\:\:Proved}\:\: \red{\bigstar}$

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$\large\bf Fundamental \: Trigonometric \\ \Large \bf Identities \\ \\ \maltese \: \: \:\sin^2\theta + \cos^2\theta=1 \\ \\ \maltese \: \: \: 1+\tan^2\theta = \sec^2\theta \\ \\\maltese \: \: \: 1+\cot^2\theta = \text{cosec}^2 \, \theta$

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Answer 2

To Prove :

[cosθ/(1-tanθ)] + [sinθ/(1-cotθ)] = cosθ + sinθ

On taking L.H.S. :

[cosθ/(1-tanθ)] + [sinθ/(1-cotθ)]

(As we know that : tanθ = sinθ/cosθ and cotθ = cosθ/sinθ)

=> [cosθ/(1-sinθ/cosθ)] + [sinθ/(1-cosθ/sinθ)]

=> [cosθ.cosθ/(cosθ-sinθ)] + [sinθ.sinθ/(sinθ-cosθ)]

=> [cos^2θ/(cosθ-sinθ)] - [sin^2θ/(cosθ-sinθ)]

=> [(cos^2θ-sin^2θ)/(cosθ-sinθ)]

[ As we know that : a^2-b^2 = (a+b)(a-b) ]

=> [(cosθ-sinθ)(cosθ+sinθ)/(cosθ-sinθ)]

=> (cosθ+sinθ) = R.H.S.

Hence Proved

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Hope it helps !!