Question

In a triangle$ABC$, if $\tan \left(\frac{A}{2}\right)=p$, $\tan \left(\frac{B}{2}\right)=p$ ,then $\frac{2(p+q)(1-pq)}{(1+p^2)(1+q^2)}$ is equal to

• A. $\sin C$
• B. $\sin 2C$
• C. $\sin A+\sin B$
• D. None of these

Answer 1

The correct option is A

$\sin C$

Explanation for correct option:

Finding the value:

Given a triangle $ABC$ we have,

$\tan \left(\frac{A}{2}\right)=p$, $\tan \left(\frac{B}{2}\right)=p$

Now let's calculate the value of an expression

$\frac{2(p+q)(1-pq)}{(1+p^2)(1+q^2)}$

Therefore,

$$\begin{gathered} \frac{2(p+q)(1-pq)}{(1+p^2)(1+q^2)}=\frac{2\left(\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)\right)\left(1-\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)\right)}{\left(1+{\tan }^2\left(\frac{A}{2}\right)\right)\left(1+{\tan }^2\left(\frac{B}{2}\right)\right)}\left[\because \tan \left(\frac{A}{2}\right)=p;\tan \left(\frac{B}{2}\right)=p\right] \\ =\frac{2\sin {\displaystyle \frac{1}{2}}\left(A+B\right)\cos {\displaystyle \frac{1}{2}}\left(A+B\right)}{\left({\sin }^2{\displaystyle \frac{A}{2}}+{\cos }^2{\displaystyle \frac{A}{2}}\right)\left({\sin }^2{\displaystyle \frac{B}{2}}+{\cos }^2{\displaystyle \frac{B}{2}}\right)} \\ =\frac{2\sin {\displaystyle \frac{1}{2}}\left(A+B\right)\cos {\displaystyle \frac{1}{2}}\left(A+B\right)}{1\times 1} \\ =\sin \left(A+B\right)\left[\because 2\sin \left(A\right)\cos \left(A\right)=\sin \left(2A\right)\right] \\ =\sin \left(\mathrm{\pi }-\mathrm{c}\right)\left[\because A+B+C=\mathrm{\pi }\right] \\ =\sin \left(\mathrm{c}\right) \end{gathered}$$

Therefore, the value of the given expression is equal to $\sin \left(\mathrm{c}\right)$

Hence, option (A) is correct.