Question

If $\alpha +\beta +\gamma =2\pi $ , then
A.$\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}+\tan \dfrac{\gamma }{2}=\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}\tan \dfrac{\gamma }{2}$
B.$\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}+\tan \dfrac{\gamma }{2}=2\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}\tan \dfrac{\gamma }{2}$
C.$\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}+\tan \dfrac{\gamma }{2}=-\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}\tan \dfrac{\gamma }{2}$
D.None of these

Answer 1

Hint: Divide the whole equation $\alpha +\beta +\gamma =2\pi $ by 2. Now, transfer $\alpha $ or $\beta $ or $\gamma $ to the other side of the equation. Now, take tan to both sides of the equation and apply the trigonometric identities, given as
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan \left( \pi -\theta \right)=-\tan \theta $

Complete step-by-step answer:
We are given $\alpha +\beta +\gamma =2\pi $ ………………………………(i)
As the given options has involvement of $\tan \dfrac{\alpha }{2},\tan \dfrac{\beta }{2},\tan \dfrac{\gamma }{2}$ it means we have to apply tan function to equation (i) with re-writing the terms (by dividing the whole equation by 2) as
$\dfrac{\alpha }{2}+\dfrac{\beta }{2}+\dfrac{\gamma }{2}=\dfrac{2\pi }{2}=\pi $
$\Rightarrow \dfrac{\alpha }{2}+\dfrac{\beta }{2}+\dfrac{\gamma }{2}=\pi -\dfrac{\gamma }{2}$ …………………………………(ii)
Now, we can take tan to both sides of the equation. So, we get the above equation as
$\tan \left( \dfrac{\alpha }{2}+\dfrac{\beta }{2} \right)=\tan \left( \pi -\dfrac{\gamma }{2} \right)$ ………………………………………(iii)
Now, as we know the trigonometric identities of $\tan \left( x+y \right)$ and $\tan \left( \pi -\theta \right)$ are given as
$\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ …………………………………..(iv)
$\tan \left( \pi -\theta \right)=-\tan \theta $ ……………………………….(v)
Now, we can use the above two equations with the equation (iii). So, simplifying LHS of equation (iii) by equation (iv) and RHS of equation (iii) by equation (v) as
$\dfrac{\tan \left( \dfrac{\alpha }{2} \right)+\tan \left( \dfrac{\beta }{2} \right)}{1-\tan \left( \dfrac{\alpha }{2} \right)\tan \left( \dfrac{\beta }{2} \right)}=-\tan \left( \dfrac{\gamma }{2} \right)$
$\Rightarrow \dfrac{\tan \left( \dfrac{\alpha }{2} \right)+\tan \left( \dfrac{\beta }{2} \right)}{1-\tan \left( \dfrac{\alpha }{2} \right)\tan \left( \dfrac{\beta }{2} \right)}=-\dfrac{\tan \left( \dfrac{\gamma }{2} \right)}{1}$
On cross multiplying the above equations, we get
$\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}=-\tan \dfrac{\gamma }{2}\left( 1-\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2} \right)$
$\Rightarrow \tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}=-\tan \dfrac{\gamma }{2}+\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}\tan \dfrac{\gamma }{2}$
$\Rightarrow \tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}+\tan \dfrac{\gamma }{2}=\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}\tan \dfrac{\gamma }{2}$
Hence, we get the above relation among $\tan \dfrac{\alpha }{2},\tan \dfrac{\beta }{2},\tan \dfrac{\gamma }{2}$ if $\alpha +\beta +\gamma =2\pi $ .
So, option (a) is the correct answer.

Note: One may go wrong if he/she directly applies tan to $\alpha +\beta +\gamma =2\pi $ to both sides. We will get a relation in $\tan \alpha ,\tan \beta ,\tan \gamma $ , which is not required. We need to get relation among $\tan \dfrac{\alpha }{2},\tan \dfrac{\beta }{2},\tan \dfrac{\gamma }{2}$ as per the given options. So, be careful with this step, otherwise we have to go longer to get the required answer. So, dividing the equation by 2 is the key point of the question.
One may get the answer from the given option by putting some value of $\alpha ,\beta ,\gamma $ whose sum is $2\pi $ . Example: - 60, 60, 240 or 90, 90, 180 or 120, 120, 120.
So, it can be another approach to get the answer.