Question

$\frac{1}{ \sec(a) - 1 } - \frac{1}{ \ \sec(a) + 1 } = 2 \csc(a) \cot(a)$ Please solve this sum for me.
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Answer 1

$\huge \red{ \mathfrak{solution}}$

Correct question .

To prove:

$\boxed{ \bold{\green{ \frac{1}{ \sec( \alpha ) - 1} + \frac{1}{ \sec( \alpha ) + 1} = 2 \csc( \alpha ) \cot( \alpha )}} }$

LHS

$\leadsto \frac{1}{ \sec( \alpha ) - 1 } + \frac{1}{ \sec( \alpha ) + 1 } \\ \\ \leadsto \: \frac{ \sec( \alpha ) + 1 + \sec( \alpha ) - 1 }{ \sec ^{2} ( \alpha ) - 1} \\ \\ \leadsto \: \frac{ 2 \sec( \alpha ) }{ {sec}^{2} \alpha - 1} \\ \\ \leadsto \: \frac{2 \sec( \alpha ) }{ \tan ^{2} ( \alpha ) } \\ \\ \leadsto \: 2 \sec( \alpha ) \cot ^{2} ( \alpha ) \\ \\ \leadsto \: 2 \times \frac{1}{ \cos( \alpha ) } \times \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \times \cot( \alpha ) \\ \\ \leadsto \: 2 \times \cancel \frac{ \cos( \alpha ) }{ \cos( \alpha ) } \times \frac{1}{ \sin( \alpha ) } \times \cot( \alpha ) \\ \\ \leadsto \: 2 \csc( \alpha ) \cot( \alpha )$

LHS=RHS

$\huge \mathfrak{proved}$

Formula used :

$\boxed{\longrightarrow \blue{ { \sec }^{2} \alpha - {tan}^{2} \alpha = 1} }\\ \\ \boxed{\longrightarrow \purple{ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) }}} \\ \\ \boxed{\longrightarrow \pink{ \frac{1}{ \sin( \alpha ) } = \csc( \alpha )}}$

Answer 2

1/sec-1+1/sec+1

=sec+1+sec-1/sec^2-1

=2sec/tan^2

=2×1/cos×cot^2

=2×1/cos×cos/sin×cot

=2csc cot