Answer

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**Hint:**First we will evaluate the left-hand of equation and then further the hand-hand side of the equation. We will then open the brackets. Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side.

**Complete step by step solution:**

We will start off by solving the left-hand side of the equation. Here,

we will be first making the denominator of the right-hand side same.

Here, we are using the trigonometric identity,

$1 - {\sin ^2}x = {\cos ^2}x$

Hence, the expression can be written as,

$

= (1 + \csc A)(1 - \sin A) \\

= 1 - \sin A + \csc A - \csc A.\sin A \\

$

But we know that $\csc A = \dfrac{1}{{\sin A}}$.

\[

= 1 - \sin A + \csc A - \csc A.\sin A \\

= 1 - \sin A + \csc A - \dfrac{1}{{\sin A}}.\sin A \\

= 1 - \sin A + \csc A - 1 \\

= \dfrac{1}{{\sin A}} - \sin A \\

= \dfrac{{1 - {{\sin }^2}A}}{{\sin A}} \\

= \dfrac{{{{\cos }^2}A}}{{\sin A}} \\

= \dfrac{{\cos A}}{{\sin A}}.\cos A \\

= \cot A.\cos A \\

\]

**Hence, LHS = RHS.**

**Additional Information:**An equation involving one or more trigonometrical ratio of an unknown angle is called a trigonometric equation. a trigonometric equation is different from a trigonometric identity. An identity is satisfied for every value of the unknown angle. The value of an unknown angle which satisfies the given equation is called the root of an equation.

**Note:**While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities ${\sin ^2}x + {\cos ^2}x = 1$. While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $\cot x = \,\dfrac{{\cos x}}{{\sin x}}$. While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity.

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