Question

If $\alpha ,\beta ,\gamma$ are the angles of the triangle then ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma$ is
A. $2$
B. $-1$
C. $-2$
D. $0$

Answer 1

Hint: To find the value of the equation ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma$we will use angle sum property and form an equation. Then using that equation, we will form another equation in terms of two angles. After taking cosine on both the sides we will solve it and make the equation like ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma$ and derive its value.
Formula used:
The trigonometric formulas are:
$\begin{align}
  & \cos (A+B)=\cos A\cos B-\sin A\sin B \\
 & {{\cos }^{2}}A=1-{{\sin }^{2}}A
\end{align}$The expansion formula is:
${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Complete step-by-step solution:
We are given a triangle having angles $\alpha ,\beta ,\gamma$ and we have to find the value of ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma$.
Using angle sum property of triangle according to which the sum of all the angles in a triangle is $\pi$, we will form an equation.
$\alpha +\beta +\gamma =\pi$
$\alpha +\beta =\pi -\gamma$
Taking cosine on both of the sides,
$\cos \left( \alpha +\beta \right)=\cos \left( \pi -\gamma \right)$
Using formula of $\cos (A+B)=\cos A\cos B-\sin A\sin B$.
$\begin{align}
  & \cos \alpha \cos \beta -\sin \alpha \sin \beta =-\cos \gamma \\
 & \cos \alpha \cos \beta +\cos \gamma =\sin \alpha \sin \beta
\end{align}$Squaring on both sides,
${{\left( \cos \alpha \cos \beta +\cos \gamma \right)}^{2}}={{\left( \sin \alpha \sin \beta \right)}^{2}}$
Using formula of expansion, we will open the brackets,
$\begin{align}
  & {{\cos }^{2}}\alpha {{\cos }^{2}}\beta +\cos {{\gamma }^{2}}+2\cos \alpha \cos \beta \cos \gamma ={{\sin }^{2}}\alpha {{\sin }^{2}}\beta \\
 & (1-{{\sin }^{2}}\alpha )(1-{{\sin }^{2}}\beta )+(1-{{\sin }^{2}}\gamma )+2\cos \alpha \cos \beta \cos \gamma ={{\sin }^{2}}\alpha {{\sin }^{2}}\beta
\end{align}$$1-{{\sin }^{2}}\beta -{{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha {{\sin }^{2}}\beta +1-{{\sin }^{2}}\gamma +2\cos \alpha \cos \beta \cos \gamma ={{\sin }^{2}}\alpha {{\sin }^{2}}\beta$
     \[\begin{align}
  & 2-({{\sin }^{2}}\beta +{{\sin }^{2}}\alpha +{{\sin }^{2}}\gamma )+2\cos \alpha \cos \beta \cos \gamma ={{\sin }^{2}}\alpha {{\sin }^{2}}\beta -{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \\
 & {{\sin }^{2}}\beta +{{\sin }^{2}}\alpha +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma =2
\end{align}\]
The value of ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma$ is $2$ when$\alpha ,\beta ,\gamma$ are the angles of the triangle. Hence the correct option is (A).
Note:
We could have also solved this question by randomly taking the values of the angles of the triangle $\alpha ,\beta ,\gamma$.
In first case let us assume the angles be $\alpha ={{30}^{0}},\beta ={{60}^{0}}$ then third angle will be $\gamma ={{90}^{0}}$. We will now substitute the values in the given equation and find the value.
 $\begin{align}
  & ={{\sin }^{2}}30+{{\sin }^{2}}60+{{\sin }^{2}}90-2\cos 30\cos 60\cos 90 \\
 & =\frac{1}{4}+\frac{3}{4}+1-2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}\times 0 \\
 & =2 \\
\end{align}$
Again we will assume another values of the angles. Let us take $\alpha ={{45}^{0}},\beta ={{45}^{0}}$ then third angle will be $\gamma ={{90}^{0}}$. Again we will substitute the values of the angles in the given equation.
     \[\begin{align}
  & ={{\sin }^{2}}45+{{\sin }^{2}}45+{{\sin }^{2}}90-2\cos 45\cos 45\cos 90 \\
 & =\frac{1}{2}+\frac{1}{2}+1-2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\times 0 \\
 & =2 \\
\end{align}\]
We can see in both the cases the value of the equation is same and exactly what we derived from another method. Hence this will be the correct answer.