If $A+B+C=\pi $, prove that ${{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)$.

Hint: For solving this question as there are squares of trigonometric ratios so, we will trigonometric formulas like formula for $\cos 2\theta $ and then further for simplifying the term written on the left-hand side we will use formulas for $\cos C+\cos D$ and $\cos \left( \pi -\theta \right)$. After that, we will prove the term on the left-hand side is equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

It is given that if $A+B+C=\pi $ and we have to prove the following equation:

${{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)$

Now, before we proceed we should know the following four formulas:

$\begin{align}

  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1..............................\left( 1 \right) \\

 & \cos 2\theta =1-2{{\sin }^{2}}\theta \\

 & \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta \\

 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}.........................\left( 2 \right) \\

 & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2}

\right)...........\left( 3 \right) \\

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C......................\left( 4 \right) \\

\end{align}$

Now, we will be using the above three formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to ${{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C$ so, using the formula from equation (1) and equation (2). Then,

$\begin{align}

  & {{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C \\

 & \Rightarrow \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+1-{{\cos }^{2}}C \\

 & \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+1-\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

 & \Rightarrow 2-\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

\end{align}$

Now, using the formula from equation (3) in the above equation. Then,

$\begin{align}

  & 2-\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

 & \Rightarrow 2-\dfrac{1}{2}\left( 2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right) \right)-{{\cos }^{2}}C \\

 & \Rightarrow 2-\cos \left( A+B \right)\cos \left( A-B \right)-{{\cos }^{2}}C \\

\end{align}$

Now, using the formula from the equation (4) in the above equation. Then,

$\begin{align}

  & 2-\cos \left( A+B \right)\cos \left( A-B \right)-{{\cos }^{2}}C \\

 & \Rightarrow 2+\cos C\cos \left( A-B \right)-{{\cos }^{2}}C \\

 & \Rightarrow 2+\cos C\left( -\cos C+\cos \left( A-B \right) \right) \\

 & \Rightarrow 2+\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\

\end{align}$

Now, using the formula from the equation (3) in the above equation. Then,

$\begin{align}

  & 2+\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\

 & \Rightarrow 2+\cos C\left( 2\cos \left( \dfrac{A+B+A-B}{2} \right)\cos \left( \dfrac{A+B-A+B}{2} \right) \right) \\

 & \Rightarrow 2+\cos C\left( 2\cos A\cos B \right) \\

 & \Rightarrow 2+2\cos A\cos B\cos C \\

 & \Rightarrow 2\left( 1+\cos A\cos B\cos C \right) \\

\end{align}$

Now, from the above result, we can say that ${{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)$.

Thus, $L.H.S=R.H.S$.

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification process smooth, we should also try to make use of trigonometric results like $\cos \left( \pi -\theta \right)=-\cos \theta $ for making equations that will help us further in the solution. Moreover, the formulas like $\cos C+\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.