Question

Solve the question in the attachment!

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Answer 1

Answer:

OPTION C IS CORRECT.

HENCE PROVED

Step-by-step explanation:

$1 + \frac{ { \cot( \alpha ) }^{2} }{1 + \csc( \alpha ) } = 1 + \frac{{ \cos( \alpha ) }^{2}}{ { \sin( \alpha ) }^{2}(1 + \csc( \alpha )) } \\ \\ = 1 + \frac{ { \cos( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} + \sin( \alpha ) } \\ \\ = \frac{ { \sin( \alpha ) }^{2} + \sin( \alpha ) + { \cos( \alpha ) }^{2} }{ { \sin( \alpha ) + \sin( \alpha ) }^{2} } \\ \\ = \frac{1 + \sin( \alpha ) }{ \sin( \alpha ) (1 + \sin( \alpha )) } \\ \\ = \frac{1}{ \sin( \alpha ) } \\ \\ = \csc( \alpha ) = rhs.$

Answer 2

Given :-

$\quad \leadsto \quad \sf 1 + \dfrac{\cot² \alpha}{1 + \cosec \alpha}$

To Find :-

Value of Given

Solution :-

We knows that ;

• $\sf \cot² \theta + 1= \cosec² \theta \: \: \forall \: \: \theta \in \mathbb R$

Using this we have ;

$\quad \leadsto \quad \bf 1 + \dfrac{\cosec² \alpha - 1}{1 + \cosec \alpha}$

further can written as ;

${ : \implies \quad \sf 1 + \dfrac{\cosec² \alpha - 1²}{1 + \cosec \alpha}}$

we knows that ;

• a² - b² = ( a + b ) ( a - b )

using this we have ;

${ : \implies \quad \sf 1 + \dfrac{( \cosec \alpha + 1 ) ( \cosec \alpha - 1 )}{1 + \cosec \alpha}}$

${ : \implies \quad \sf 1 + \dfrac{\cancel{( \cosec \alpha + 1 )} ( \cosec \alpha - 1 )}{\cancel{(cosec \alpha+1)}}}$

${ : \implies \quad \sf 1 + \cosec \alpha - 1 }$

${ : \implies \quad \sf \cancel{1} + \cosec \alpha - \cancel{1} }$

${ : \implies \quad \bf \cosec \alpha}$

Henceforth , Option (c) is correct :D

$\quad \qquad { \bigstar { \underline { \boxed { \pmb { \red { \underbrace { \bf { \therefore 1 + \dfrac{\cot² \alpha}{1 + \cosec \alpha} = \cosec \alpha }}}}}}}}{\bigstar} \quad \qquad$