Question

$\Large \frac{ \tan( \alpha ) }{1 - \cot( \alpha ) } + \frac{ \cot( \alpha ) }{1 - \tan( \alpha ) }$ is Equal to which of the following Option :

$1)1 - \tan( \alpha ) - \cot( \alpha )$
$2)1 + \tan( \alpha ) - \cot( \alpha )$
$3)1 - \tan( \alpha ) + \cot( \alpha )$
$4)1 + \tan( \alpha ) + \cot( \alpha )$
​

Answer 1

AnswEr :

$\longrightarrow\dfrac{ \tan( \alpha ) }{1 - \cot( \alpha ) } + \dfrac{ \cot( \alpha ) }{1 - \tan( \alpha )}$

⠀

• we will change all cot α into tan α

⠀

$\longrightarrow\dfrac{ \tan( \alpha ) }{1 - \frac{1}{ \tan( \alpha ) } } + \dfrac{ \frac{1}{ \tan( \alpha ) } }{1 - \tan( \alpha )}$

$\longrightarrow\dfrac{ \tan( \alpha ) }{\frac{ \tan( \alpha) - 1}{ \tan( \alpha ) } } + \dfrac{ \frac{1}{ \tan( \alpha ) } }{1 - \tan( \alpha )}$

$\longrightarrow\dfrac{ \tan( \alpha ) \times \tan( \alpha ) }{\tan( \alpha) - 1}+ \bigg(\dfrac{1}{ \tan( \alpha ) } \times \dfrac{1}{{1 - \tan( \alpha )}} \bigg)$

$\longrightarrow\dfrac{ \tan^{2} ( \alpha ) }{\tan( \alpha) - 1}+ \dfrac{1}{ \tan( \alpha ) - { \tan}^{2} (\alpha )}$

⠀

• taking Negative Common

⠀

$\longrightarrow\dfrac{ \tan^{2} ( \alpha ) }{\tan( \alpha) - 1}+ \dfrac{1}{ - (-\tan( \alpha ) + { \tan}^{2} (\alpha))}$

$\longrightarrow\dfrac{ \tan^{2} ( \alpha ) }{\tan( \alpha) - 1} - \dfrac{1}{( { \tan}^{2} (\alpha ) - \tan( \alpha ) )}$

$\longrightarrow\dfrac{1}{\tan( \alpha ) - 1} \times\bigg( \tan^{2} ( \alpha ) - \dfrac{1}{ \tan( \alpha )}\bigg)$

$\longrightarrow\dfrac{1}{\tan( \alpha ) - 1} \times\bigg( \dfrac{ { \tan }^{3}( \alpha ) - 1}{ \tan( \alpha )}\bigg)$

$\longrightarrow\dfrac{1}{\tan( \alpha ) - 1} \times\bigg( \dfrac{ { \tan }^{3}( \alpha ) - {1}^{3} }{ \tan( \alpha )}\bigg)$

⠀

• a³ - b³ = (a - b)(a² + ab + b²)

⠀

$\longrightarrow\dfrac{1}{ \cancel{\tan( \alpha ) - 1}} \times\dfrac{ \cancel{(\tan( \alpha ) - 1)}( \tan ^{2} ( \alpha ) + {1}^{2} + 1 \times \tan( \alpha )) }{ \tan( \alpha )}$

$\longrightarrow \dfrac{ \tan^{2} ( \alpha ) + 1 + \tan( \alpha ) }{ \tan( \alpha ) }$

$\longrightarrow \cancel\dfrac{ \tan^{2} ( \alpha )}{ \tan( \alpha ) } + \dfrac{ 1}{ \tan( \alpha ) } + \cancel\dfrac{ \tan ( \alpha )}{ \tan( \alpha ) }$

$\longrightarrow \tan( \alpha ) + \cot( \alpha ) + 1$

$\longrightarrow 1 + \tan( \alpha ) + \cot( \alpha )$

⠀

$\therefore\underline{\sf{Correct\: Option\:is\:4)1 + \tan( \alpha ) + \cot( \alpha ) }}$

Answer 2

Answer:

Refer to the attachment for Solution.

Please Mark Brainliest ❤️