Question

If $\alpha + \beta + \gamma = \pi $, then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma $ is equal to
A.$2\sin \alpha \sin \beta \cos \gamma $
B.$2\cos \alpha \cos \beta \cos \gamma $
C.$2\sin \alpha \sin \beta \sin \gamma $
D.None of the above

Answer 1

Hint: In order to solve the given question, we will simplify the given angle in terms of $\alpha $ and $\beta $ . Then we will put some fundamental formulas of trigonometry. After all of that we will do further simplification and put our angle values and find out our final answers. Some special formulas are used in this question are,
${\sin ^2}A = 1 - {\cos ^2}A$
${\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}$

Complete answer:
First of all, we have given $\alpha + \beta + \gamma = \pi $, we can write this term also in this way $\alpha + \beta = \pi - \gamma $. We can multiply it by 2 and we get $2\alpha + 2\beta = 2\pi - 2\gamma $.
Now, Our given question is
$ \Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma $
Now, replace some terms with following formula,
${\sin ^2}A = 1 - {\cos ^2}A$
And
 ${\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}$.
After replacement we get,
$ \Rightarrow (\dfrac{{1 - \cos 2\alpha }}{2}) + (\dfrac{{1 - \cos 2\beta }}{2}) - (1 - {\cos ^2}\gamma )$
Take $ - (\dfrac{1}{2})$ common and implement this formula $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$
$ \Rightarrow 1 - \dfrac{1}{2}(\cos 2\alpha + \cos 2\beta ) - (1 - {\cos ^2}\gamma )$
After some for simplification, we get
$ \Rightarrow - \dfrac{1}{2}(2\cos (\alpha + \beta )\cos (\alpha - \beta )) + {\cos ^2}\gamma $
But as we know $\alpha + \beta = \pi - \gamma $and $\cos (\pi + A) = - \cos A$ so we can do further steps,
$ \Rightarrow - \cos (\pi + \gamma )\cos (\alpha - \beta ) + {\cos ^2}\gamma $
On further simplification we get,
$ \Rightarrow \cos \gamma \cos (\alpha - \beta ) + {\cos ^2}\gamma $
Now, let’s come to our final step. In this step we will take $\cos \gamma $ common and implement $\cos A - \cos B = 2\sin (\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2})$ this formula we will get our answer.
$ \Rightarrow \cos \gamma (\cos (\alpha - \beta ) + \cos (\alpha + \beta - \pi )$
we know that ($\cos ( - A) = \cos A$)
$ \Rightarrow \cos \gamma (\cos (\alpha - \beta ) + \cos (\pi - (\alpha + \beta ))$
Do some more calculation and we get,
$ \Rightarrow \cos \gamma (\cos (\alpha - \beta ) - \cos (\alpha + \beta ))$
$ \Rightarrow \cos \gamma (2\sin (\dfrac{{\alpha - \beta + \alpha + \beta }}{2})\sin (\dfrac{{\alpha + \beta - (\alpha - \beta )}}{2}))$
Just cancel out bracket things and we get,
$ \Rightarrow 2\sin \alpha \sin \beta \cos \gamma $.
Hence, ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma \, = \,2\sin \alpha \sin \beta \cos \gamma $ .
Therefore the correct answer is option A .

Note:
The sum of the squares of sine and cosine angle is equal to 1. It is an identity so it can be used everywhere or to solve these types of problems but there are some certain formulae that are valid at some definite interval. So in order to solve these types of problems one should check the formulae and their intervals where they hold true.