Trigonometry

Prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

To prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

i.e. $\;$ To prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} + \dfrac{1}{\text{cosec }A + \cot A} = \dfrac{1}{\sin A} + \dfrac{1}{\sin A}$

i.e. $\;$ To prove that $\;$ $\dfrac{\text{cosec }A + \cot A + \text{cosec }A - \cot A}{\left(\text{cosec }A + \cot A\right) \left(\text{cosec }A - \cot A\right)} = \dfrac{2}{\sin A}$

i.e. $\;$ To prove that $\;$ $\dfrac{2 \; \text{cosec }A}{\text{cosec}^2 A - \cot^2 A} = 2 \; \text{cosec }A$

Now, $\;$ $1 + \cot^2 A = \text{cosec}^2 A$

$\implies$ $\text{cosec}^2 A - \cot^2 A = 1$

$\therefore \;$ To prove that $\;$ $2 \; \text{cosec }A = 2 \; \text{cosec A}$ $\;\;$ which is true.

$\therefore \;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

Hence proved.