Question

If $\alpha +\beta =\dfrac{\pi }{2}$ and $\beta +\gamma =\alpha $, then $\tan \alpha $ equals:
(a) $2\left( \tan \beta +\tan \gamma \right)$
(b) $\tan \beta +\tan \gamma $
(c) $\tan \beta +2\tan \gamma $
(d) $2\tan \beta +\tan \gamma $

Answer 1

Hint: We have given the equation $\alpha +\beta =\dfrac{\pi }{2}$, subtracting $\beta $ on both the sides and then take tan on both the sides will give the equation $\tan \alpha =\tan \left( \dfrac{\pi }{2}-\beta \right)$ further solving this equation will give you a relation between $\tan \alpha \And \tan \beta $. Mark it as eq. (1) then solve the other equation i.e. $\beta +\gamma =\alpha $ by subtracting $\beta $ on both the sides after that take tan on both the sides of the equation then simplify it further and use the eq. (1) to simplify it further and will get you a relation between $\tan \alpha ,\tan \beta \And \tan \gamma $

Complete step by step answer:
We have given the following equation,
$\alpha +\beta =\dfrac{\pi }{2}$
Subtracting $\beta $ on both the sides we get,
$\alpha =\dfrac{\pi }{2}-\beta $
Taking tan on both the sides of the above equation we get,
$\tan \alpha =\tan \left( \dfrac{\pi }{2}-\beta \right)$
We know from the trigonometric ratios that $\tan \left( \dfrac{\pi }{2}-\beta \right)=\cot \beta $ using this relation in the above expression we get,
$\tan \alpha =\cot \beta $
Using the trigonometric relation of $\cot \beta =\dfrac{1}{\tan \beta }$ in the above equation we get,
$\tan \alpha =\dfrac{1}{\tan \beta }$
Cross multiplying the above equation we get,
$\tan \alpha \left( \tan \beta \right)=1$……….. Eq. (1)
We have also given another equation as follows:
$\beta +\gamma =\alpha $
Subtracting $\beta $ on both the sides of the above equation we get,
$\gamma =\alpha -\beta $
Now, taking tan on both the sides of the above equation we get,
$\tan \gamma =\tan \left( \alpha -\beta \right)$
We know the identity of $\tan \left( \alpha -\beta \right)=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$ using this identity in the above equation we get,
$\tan \gamma =\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$
Using eq. (1) we can substitute the value of $\tan \alpha \tan \beta =1$ in the above equation we get,
$\begin{align}
  & \tan \gamma =\dfrac{\tan \alpha -\tan \beta }{1+1} \\
 & \Rightarrow \tan \gamma =\dfrac{\tan \alpha -\tan \beta }{2} \\
\end{align}$
Cross multiplying the above equation we get,
$2\tan \gamma =\tan \alpha -\tan \beta $
Adding $\tan \beta $ on both the sides of the above equation we get,
$2\tan \gamma +\tan \beta =\tan \alpha $
Now, compare the relation that we are getting with the options given in the question.

So, the correct answer is “Option c”.

Note: You might have thought that in simplifying the equation $\beta +\gamma =\alpha $, why we have subtracted $\beta $ on both the sides and then take tan on both the sides instead we can subtract $\gamma $ and then take tan on both the sides also.
The answer is let us subtract $\gamma $ on both the sides of the equation $\beta +\gamma =\alpha $ we get,
$\beta =\alpha -\gamma $
Taking tan on both the sides of the above equation we get,
$\begin{align}
  & \tan \beta =\tan \left( \alpha -\gamma \right) \\
 & \Rightarrow \tan \beta =\dfrac{\tan \alpha -\tan \gamma }{1+\tan \alpha \tan \gamma } \\
\end{align}$
Now, you can see that we cannot use the relation in eq. (1) i.e. $\tan \alpha \left( \tan \beta \right)=1$ this is the reason why we haven’t simplified the equation in this way.