Question

Prove that : COS A /1-tan A+ sin A/ 1-cot A =sin A+cos A​

Answer 1

Question:-

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

To Prove:-

$L.H.S=R.H.S$

Solution:-

Solving L.H.S

$L.H.S = \frac{ \cos( \alpha ) }{1 - \tan( \alpha ) } + \frac{ \sin( \alpha ) }{1 - \cot( \alpha ) }$

Now We Will Solve the denominator

we know that :-

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

and

$\cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) }$

so,

we can also write L.H.S as

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

Now

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

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

$= > \frac{ \cos( \alpha ) \times \cos( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } + \frac{ \sin( \alpha ) \times \sin( \alpha ) }{ \sin( \alpha ) - \cos( \alpha ) }$

$= > \frac{ { \cos( \alpha ) }^{2} }{ \cos( \alpha ) - \sin( \alpha ) } + \frac{ { \sin( \alpha ) }^{2} }{ \sin( \alpha ) - \cos( \alpha ) }$

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

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

we can also write numerator as

${ \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} = ( \cos( \alpha ) + \sin( \alpha ) )( \cos( \alpha ) - \sin( \alpha ) )$

Because of third Identity

i.e,

${x}^{2} - {y}^{2} = (x - y)(x + y)$

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

$= >\frac{( \cos( \alpha )+ \sin( \alpha ) ) \cancel{ ( \cos( \alpha ) - \sin( \alpha ) )}}{ \cancel{\cos( \alpha ) - \sin( \alpha ) }}$

$= > \cos( \alpha )+ \sin( \alpha )$

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

$\therefore \: L.H.S=R.H.S$

HENCE PROVED

Answer 2

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