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**Hint:**In this question, we will proceed by considering the L.H.S part of the given equation. Then use the formula in trigonometric identities and trigonometric ratios to prove that the L.H.S part of the given equation is equal to the R.H.S part.

**Complete step-by-step answer:**

Given equation is $\operatorname{cosec} \theta \sqrt {1 - {{\cos }^2}\theta } = 1$

Consider the L.H.S part of the equation

We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $. By using this formula, we have

$

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}} = \operatorname{cosec} \theta \sqrt {{{\sin }^2}\theta } \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}} = \operatorname{cosec} \theta \left( {\sin \theta } \right) \\

$

We know that $\operatorname{cosec} \theta = \dfrac{1}{{\sin \theta }}$. By using this formula, we have

$

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}} = \operatorname{cosec} \theta \sqrt {{{\sin }^2}\theta } \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}} = \dfrac{1}{{\sin \theta }}\left( {\sin \theta } \right) = \dfrac{{\sin \theta }}{{\sin \theta }} \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}} = 1 \\

\therefore {\text{L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S}} \\

$

Hence, proved that $\operatorname{cosec} \theta \sqrt {1 - {{\cos }^2}\theta } = 1$.

**Note:**Here we have used the trigonometry identity ${\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow 1 - {\cos ^2}\theta = {\sin ^2}\theta $ and the trigonometric ratio $\operatorname{cosec} \theta = \dfrac{1}{{\sin \theta }}$. So, in solving these types of questions, remember all the formula n trigonometry to solve easily.

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