Question

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

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Answer 1

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$\frac{cosΦ}{1\frac{sinΦ}{cosΦ} } + \frac{sinΦ}{1 - \frac{cosΦ}{sinΦ} }$

$\frac{cosΦ}{ \frac{cosΦ - sinΦ}{cosΦ} } + \frac{sinΦ}{ \frac{sinΦ - cosΦ}{sinΦ} }$

$\frac{ {cos}^{2Φ} }{cosΦ - sinΦ} + \frac{ {sin}^{2Φ} }{sinΦ - cosΦ}$

$- \frac{ {cos}^{2Φ} }{cosΦ - sinΦ} + \frac{ {sin}^{2Φ} }{cosΦ - sinΦ}$

$\frac{ {sin }^{2Φ - {cos}^{2Φ} } }{cosΦ - sinΦ}$

$\frac{(sinΦ + cosΦ)(sinΦ - cosΦ)}{sinΦ -cosΦ }$

➪sinΦ+cosΦ answer.....

$\huge\{\boxed{\colorbox{pink}{Hope it'z help you ♡︎}}$

Answer 2

Solution:-

$\frac{ \frac{ \cosθ }{1 - \sinθ } }{ \cosθ } + \frac{ \sinθ }{ \frac{1 - \cosθ }{ \sinθ } }$

$\frac{ \cos {}^{2} θ }{ \cosθ - \sinθ } + \frac{ \sin {}^{2} {θ}^{} }{ \sinθ - \cosθ }$

$\frac{ \cos {}^{2} {θ}^{} - \sin {}^{2} θ }{ \cosθ - \sinθ }$

$\frac{ (\cosθ - \sinθ )( \cosθ + \sinθ }{ \cosθ - \sinθ }$

$= ( \cosθ + \sinθ)$

●Hope it helps you.