Question

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Answer 1

Question

$\rm \to \dfrac{ \cos x }{1 - \tan x } - \dfrac{ \sin {}^{2} x }{ \cos x - \sin x } = \cos x + \sin x$

Solution:-

$\rm \to \dfrac{ \cos x }{1 - \tan x} - \dfrac{ \sin {}^{2} x }{ \cos x - \sin x } = \cos x + \sin x$

We have to prove

$\rm \cos x + \sin x$

Take LHS

$\rm \to \dfrac{ \cos x }{1 - \tan x } - \dfrac{ \sin {}^{2} x }{ \cos x - \sin x}$

Use this trigonometry identities

$\rm \: \implies \tan x = \dfrac{ \sin x}{ \cos x}$

We get

$\rm \to \dfrac{ \cos x }{1 - \dfrac{ \sin x}{ \cos x} } - \dfrac{ \sin {}^{2} x }{ \cos x- \sin x }$

take lcm

$\rm \to \dfrac{ \cos x }{ \dfrac{ \cos x - \sin x}{ \cos x } } - \dfrac{ \sin {}^{2} x}{ \cos x- \sin x }$

$\rm \to \dfrac{ \cos {}^{2} x }{ \cos x - \sin x } - \dfrac{ \sin {}^{2} x }{ \cos x - \sin x }$

Take lcm

$\rm\dfrac{ \cos {}^{2} x - \sin {}^{2} x }{ \cos x - \sin x }$

use this identity;

$\rm( {a}^{2} - {b}^{2} ) = ( a + b)(a - b)$

we get

$\rm\dfrac{( \cos {}^{} x - \sin {}^{} x ) ( \cos x + \sin x )}{ \cos x - \sin x}$

Now

$\rm \: \cos x - \sin x \: \: is \: cancel \: out$

we get

$\rm \cos x + \sin x$

LHS = RHS , Hence proved