Question

Pls help me to solve the problem ​

Answer 1

EXPLANATION.

$\sf : \implies \: \: \dfrac{ \tan(a) }{ \sec(a) - 1 } + \dfrac{ \tan(a) }{ \sec(a) + 1} = 2 \csc(a)$

$\sf : \implies \: \frac{ \dfrac{ \sin(a) }{ \cos(a) } }{ \dfrac{1}{ \cos(a) } - 1 } \: \: + \: \: \frac{ \dfrac{ \sin(a) }{ \cos(a) } }{ \dfrac{1}{ \cos(a) } + 1}$

$\sf : \implies \: \dfrac{ \dfrac{ \sin(a) }{ \cos(a) } }{ \dfrac{1 - \cos(a) }{ \cos(a) } } \: \: + \: \: \: \dfrac{ \dfrac{ \sin(a) }{ \cos(a) } }{ \dfrac{1 + \cos(a) }{ \cos(a) } }$

$\sf : \implies \: \dfrac{ \sin(a) }{ \cancel{\cos(a)} } \times \dfrac{ \cancel{\cos(a)} }{1 - \cos(a) } \: \: + \: \: \dfrac{ \sin(a) }{ \cancel{\cos(a)} } \times \dfrac{ \cancel{\cos(a)} }{1 + \cos(a) }$

$\sf : \implies \: \dfrac{ \sin(a) }{1 - \cos(a) } \: \: + \: \: \: \dfrac{ \sin(a) }{1 + \cos(a) }$

$\sf : \implies \: \dfrac{ \sin(a) (1 + \cos(a) ) + \sin(a) (1 - \cos(a) )}{(1 - \cos(a) )(1 + \cos(a)) }$

$\sf : \implies \: \dfrac{ \sin(a) (1 + \cos(a) + 1 - \cos(a) )}{(1 - \cos {}^{2} (a) )}$

$\sf : \implies \: \dfrac{ 2\sin(a) }{ \sin {}^{2} (a) } = \dfrac{2}{ \sin(a) } = 2 \csc(a)$

$\sf : \implies \: \green{{ \underline{formula \: used \: }}} \\ \\ \sf : \implies \: \tan(a) = \frac{ \sin(a) }{ \cos(a) } \\ \\ \sf : \implies \: \sec(a) = \frac{1}{ \cos(a) } \\ \\ \sf : \implies \: \csc(a) = \frac{1}{ \sin(a) } \\ \\ \sf : \implies \: 1 - \cos {}^{2} (a) = \sin {}^{2} (a)$

Answer 2

To Prove:-

$\rm\dfrac{\tan A}{\sec A -1} + \dfrac{\tan A}{ \sec A +1} = 2 \ cosec \ A$

Solution:-

$\rm\dfrac{\tan A}{\sec A -1} + \dfrac{\tan A}{ \sec A +1} = 2 \ cosec \ A$

Take $\tan A$ common.

$\rm\dashrightarrow \tan A \Bigg ( \dfrac{1}{\sec A -1} + \dfrac{1}{ \sec A +1} \Bigg )= 2 \ cosec \ A$

$\rm\dashrightarrow \tan A \Bigg ( \dfrac{\sec A +1 + (\sec A - 1 )}{ (\sec A -1 ) ( \sec A +1 ) } \Bigg )= 2 \ cosec \ A$

$\rm\dashrightarrow \tan A \Bigg ( \dfrac{ 2 \sec A}{{ \sec }^2 A-1} \Bigg )= 2 \ cosec \ A \qquad\qquad \bigg [ \because \ \ a^2-b^2 = (a-b)(a+b) \bigg ]$

$\rm\dashrightarrow \tan A \Bigg ( \dfrac{ 2 \sec A}{{\tan}^2 A} \Bigg )= 2 \ cosec \ A \qquad \qquad ...\Bigg [ \because \ \ {\sec}^2 A -1 = {\tan}^2 A \Bigg ]$

$\rm\dashrightarrow \dfrac{ 2 \sec A}{\tan A} = 2 \ cosec \ A$

$\rm\dashrightarrow \dfrac{ \frac{2}{\cos A}}{\frac{\sin A}{\cos A}} = 2 \ cosec \ A$

$\dashrightarrow\rm \dfrac{2}{\sin A} = 2 \ cosec \ A$

$\dashrightarrow \rm 2 \ cosec \ A= 2 \ cosec \ A$