Prove the following identities :

$\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A

Multiplying numerator and denominator by $\sqrt{1 - \text{cos A}}$

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

By formula,

1 - cos² A = sin² A

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

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

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