Question

please solve it step by step ​

Answer 1

Answer:

LHS =

$\frac{tan \: a}{sec \: a - 1} + \frac{tan \: a}{sec \: a + 1}$

$= tan \: a( \frac{sec \: a + 1 + sec \: a - 1}{ {sec}^{2} a - 1} )$

$= \frac{tan \: a \times 2sec \: a}{ {tan}^{2}a }$

$= 2cosec \: a$

= RHS

Hence, Proved...

Answer 2

Step-by-step explanation:

${ \sf{ \underline{ \red{ \underline{ \red{Given}}}} : - }}$

$\displaystyle \bigstar \: \: \: \: \boxed{ \underline{ \boxed{ \bold{ \frac{ \tan(a) }{ \sec(a - 1) } + \ \frac{ \tan(a) }{ \sec(a + 1) } }}}}$

${ \sf{ \underline{ \red{ \underline{ \red{Solution}}}} : - }}$

$\displaystyle \longrightarrow {\sf{\rm{ \green{\frac{ \tan(a) }{ \sec(a - 1) } + \frac{ \tan(a) }{ \sec(a + 1) } }}}} \: \: \\ \\ \displaystyle \longrightarrow {\sf{\rm{ \green{\frac{ \frac{ \sin(a) }{ \cos(a) } }{ \frac{1}{ \cos(a) - 1} } + \frac{ \frac{ \sin(a) }{ \cos(a) } }{ \frac{1}{ \cos(a) + 1 } }}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \: {\sf{\rm{ \green{ \frac{ \sin(a) }{1 - \cos(a) } + \frac{ \sin(a) }{1 + \cos(a)) }}}}} \\ \\ \displaystyle \longrightarrow \: {\sf{\rm{ \green{ \frac{ \sin(a)(1 + \cos(a) + \sin(a) (1 - \cos(a)}{(1 + \cos(a)(1 - \cos(a) ) } }}}} \\ \\ \displaystyle \longrightarrow \: {\sf{\rm{ \green{ \frac{2 \sin(a) }{ { \sin }^{2}a } }}}} \\ \\ \displaystyle \longrightarrow \: {\sf{\rm{ \green{2 \cosec(a) }}}}$