Question

sorry for last time !!!
see the attachment and answer asap please​

Answer 1

To prove :

$\star \: \: \dfrac{1 + \cos( \alpha ) }{ \sin( \alpha ) } + \dfrac{ \sin( \alpha ) }{1 + \cos( \alpha ) } = 2 \csc( \alpha )$

LHS :

$\implies \: \dfrac{1 + \cos( \alpha ) }{ \sin( \alpha ) } + \dfrac{ \sin( \alpha ) }{1 + \cos( \alpha ) } \\ \\ \implies \: \frac{( {1 + \cos( \alpha )) }^{2} + { \sin}^{2} \alpha }{ \sin( \alpha )(1 + \cos( \alpha )) } \\ \\ \implies \: \frac{1 + { \cos}^{2} \alpha + 2 \cos( \alpha ) + { \sin}^{2} \alpha }{ \sin( \alpha ) (1 + \cos( \alpha )) } \\ \\ \implies \: \frac{1 + 1 + 2 \cos( \alpha ) }{ \sin( \alpha )(1 + \cos( \alpha )) } \\ \\ \implies \: \frac{2 + 2 \cos( \alpha ) }{ \sin( \alpha )(1 + \cos( \alpha )) } \\ \\ \implies \: \frac{2( \cancel{1 + \cos( \alpha )}) }{ \sin( \alpha ) \cancel{(1 + \cos( \alpha )) }} \\ \\ \implies \: \frac{2}{ \sin( \alpha ) } \\ \\ \implies \: 2 \csc( \alpha )$

LHS = RHS .

Proved .

Formula used :

$\star \: { \sin}^{2} \theta + { \cos }^{2} \theta = 1 \\ \\ \star \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ \star \: \sin \theta = \frac{1}{ \cosec \theta}$

Answer 2

Answer:

HENCE IT PROVED

Step-by-step explanation:

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