If $\left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)$ , then each side is equal to ?

Hint: Here, side does not refer to any polygon. The question just implies the left hand side and the right hand side. For these kinds of questions, we need to use some trigonometric formulae and a lot of manipulation is required. We will first covert $\left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)$ in terms of $\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)$ using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . And do some manipulations to arrive at the answer.

Complete step by step solution:
First we are given $\left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)......eqn\left( 1 \right)$ .
But we already know that
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
\end{align}\]
We know the formula of ${{a}^{2}}-{{b}^{2}}$ . It is ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ . Let us apply the formula and expand.
Upon doing so , we get the following :
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
 & \Rightarrow {{\cos }^{2}}\theta =\left( 1-\sin \theta \right)\left( 1+\sin \theta \right) \\
 & \Rightarrow \left( 1+\sin \theta \right)=\dfrac{{{\cos }^{2}}\theta }{\left( 1-\sin \theta \right)}....eqn\left( 2 \right) \\
\end{align}\]
Let us name it $eqn\left( 2 \right)$ .
Now let's use $eqn\left( 2 \right)$ , let us rewrite $eqn\left( 1 \right)$ .
Upon doing, we get the following :
$\Rightarrow \left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\dfrac{{{\cos }^{2}}A}{\left( 1-\sin A \right)}\dfrac{{{\cos }^{2}}B}{\left( 1-\sin B \right)}\dfrac{{{\cos }^{2}}C}{\left( 1-\sin C \right)}$
\[\begin{align}
 & \Rightarrow \left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\dfrac{{{\cos }^{2}}A{{\cos }^{2}}B{{\cos }^{2}}C}{\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)} \\
\end{align}\]
Let us cross-multiply.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow {{\left[ \left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right) \right]}^{2}}={{\cos }^{2}}A{{\cos }^{2}}B{{\cos }^{2}}C \\
 & \Rightarrow \left[ \left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right) \right]=\sqrt{{{\cos }^{2}}A{{\cos }^{2}}B{{\cos }^{2}}C} \\
 & \Rightarrow \left[ \left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right) \right]=\pm \cos A\cos B\cos C \\
\end{align}\]
Since $\left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)$, we can conclude the following :
$\Rightarrow $ \[\left[ \left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right) \right]=\pm \cos A\cos B\cos C\] .

$\therefore $ Hence, If $\left( 1+\sin A \right)\left( 1+\sin B \right)\left( 1+\sin C \right)=\left( 1-\sin A \right)\left( 1-\sin B \right)\left( 1-\sin C \right)$ , then each side is equal to \[\pm \cos A\cos B\cos C\].

Note: It is important all the trigonometric formulae, definitions to solve the questions quickly. We should be very careful while solving as there is a lot of scope for calculation errors. Trigonometry is a kind of chapter where there is involvement of many manipulations to arrive at the result. So a lot of practice must be put in to solve a question quickly and confidently in the exam. It is also important to remember all the formulae of properties of triangles.