Question

♧ Bʀᴀɪɴʟʏ Sᴛᴀʀs

♧ Mᴏᴅᴇʀᴀᴛᴇʀs

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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

To Prove :

LHS = RHS.

SolutioN :

$\tt{ \dfrac{ Cos \theta}{1 - tan \theta} + \dfrac{ Sin \theta}{1 - Cot \theta} = Cos \theta + Sin \theta}$

⚝ Taking LHS

$\rightarrow \tt \dfrac{ Cos \theta}{1 - tan \theta} + \dfrac{ Sin \theta}{1 - Cot \theta}$

$\rightarrow \tt \dfrac{ Cos \theta}{1 - \frac{Sin\,\theta}{Cos\,\theta} } + \dfrac{ Sin\, \theta}{1 - \frac{Cos\, \theta}{Sin\,\theta}}$

$\rightarrow \tt \dfrac{ Cos \theta}{ \frac{Cos\,\theta-Sin\,\theta}{Cos\,\theta} } + \dfrac{ Sin\, \theta}{ \frac{Sin\,\theta - Cos\, \theta}{Sin\,\theta}}$

⚝ Taking sign ( - ) common.

$\rightarrow \tt \dfrac{ Cos^2 \theta- Sin^2\,\theta}{Cos\,\theta-Sin\,\theta}$

$\rightarrow \tt \dfrac{ (Cos\theta- Sin\,\theta)( Cos\theta + Sin\,\theta )}{Cos\,\theta-Sin\,\theta}$

$\rightarrow \tt Cos\theta + Sin\,\theta$

Hence Proved.

Answer 2

$\small\pink{Hope \: you \: have \: satisfied.}$