Question

prove that cos (therta) /1-tan(therta)+sin(therta)1-cot(therta)=(cos(therta)+sin(therta)​

Answer 1

Question

$\frac{ \cos(x) }{1 - \tan(x) } + \frac{ \sin(x) }{1 - \cot(x) } = \cos(x) + \sin(x)$

Solution :

$\frac{ \cos(x) }{1 - \tan(x) } + \frac{ \sin(x) }{1 - \cot(x) } = \cos(x) + \sin(x)$

LHS :

$\frac{ \cos(x) }{1 - \tan(x) } + \frac{ \sin(x) }{1 - \cot(x) }$

convert tanx and cotx in terms of sinx and cosx

$= \frac{ \cos(x) }{1 - \frac{ \sin(x) }{ \cos(x) } } + \frac{ \sin(x) }{1 - \frac{ \cos(x) }{ \sin(x) } }$

$= \frac{ \cos(x) }{ \frac{ \cos(x) - \sin(x) }{ \cos(x) } } + \frac{ \sin(x) }{ \frac{ \sin(x) - \cos(x) }{ \sin(x) } }$

$= \frac{ \cos {}^{2} (x) }{ \cos(x) - \sin(x) } - \frac{ \sin {}^{2} (x) }{ \cos(x) - \sin(x) }$

$= \frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{ \cos(x) - \sin(x) }$

use formula

a²- b² = (a+b) (a-b)

$= \frac{( \cos(x) + \sin(x))( \cos(x) - \sin(x) )}{ \cos(x) - \sin(x) }$

$= \cos(x) + \sin(x)$

$\huge{\bold{ Hence \: Proved }}$

Answer 2

❥${\underline{\underline{\huge{\mathbb{QUESTION:-}}}}}$

Prove that,

${\bold{\frac{cos\theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}=cos\theta+sin\theta}}$

❥${\underline{\underline{\huge{\mathbb{SOLUTION:-}}}}}$

† Taking L.H.S †

${\blue{\bold{\frac{cos\theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}}}}$

★We know ${\bold{\tan\theta=\frac{sin\theta}{cos\theta}}}$ and ${\bold{\cot\theta=\frac{cos\theta}{sin\theta}}}$★

${\bold{→\frac{cos\theta}{1-\frac{sin\theta}{cos\theta}}}+\frac{sin\theta}{1-\frac{cos\theta}{sin\theta}}}$

${\bold{→\frac{cos\theta}{\frac{cos\theta-sin\theta}{cos\theta}}}}$+${\bold{\frac{sin\theta}{\frac{sin\theta-cos\theta}{sin\theta}}}}$

${\bold{→cos\theta×\frac{cos\theta}{cos\theta-sin\theta}+sin\theta×\frac{sin\theta}{sin\theta-cos\theta}}}$

${\bold{→\frac{cos^2\theta}{cos\theta-sin\theta}+\frac{sin^2\theta}{sin\theta-cos\theta}}}$

${\bold{→\frac{cos^2\theta}{cos\theta-sin\theta}-\frac{sin^2\theta}{cos\theta-sin\theta}}}$

${\bold{→\frac{cos^2\theta-sin^2\theta}{cos\theta-sin\theta}}}$

★Apply a²-b²=(a+b)(a-b)★

${\bold{→\frac{(cos\theta+sin\theta)(cos\theta-sin\theta)}{cos\theta-sin\theta}}}$

${\red{\bold{→cos\theta+sin\theta}}}$

L.H.S = R.H.S

Hence proved________!!