Prove that :

$\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}}$ = 2 tan A

To prove:

$\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}}$ = 2 tan A

Solving L.H.S. of the equation

$\Rightarrow \dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} \\[1em] \Rightarrow \dfrac{\text{cos A(cosec A - 1) + cos A(cosec A + 1)}}{\text{(cosec A + 1)(cosec A - 1)}} \\[1em] \Rightarrow \dfrac{\text{cos A.cosec A - cos A + cos A.cosec A+ cos A}}{\text{cosec}^2 A - 1}$

By formula,

cosec² A - 1 = cot² A

$\Rightarrow \dfrac{\text{2 cos A cosec A}}{\text{cot}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos A} \times \dfrac{1}{\text{sin A}}}{\text{cot}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cot A}}{\text{cot}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{cot A}} \\[1em] \Rightarrow \text{2 tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}}$ = 2 tan A.