Question

If $\sin \left( {\alpha + \beta } \right) = 1,\sin \left( {\alpha - \beta } \right) = 1/2$, then $\tan \left( {\alpha + 2\beta } \right)\tan \left( {2\alpha + \beta } \right) = $
$\left( a \right){\text{ 1}}$
$\left( b \right){\text{ - 1}}$
$\left( c \right){\text{ 0}}$
$\left( d \right){\text{ None}}$

Answer 1

Hint:
To solve this question we have to first calculate the $\alpha \& \beta $ and for this, by using the equation from the $\sin \left( {\alpha + \beta } \right) = 1,\sin \left( {\alpha - \beta } \right) = 1/2$ , from its interval we will get the value of $\alpha \& \beta $ . And then we will calculate the value $\tan \left( {\alpha + 2\beta } \right)\tan \left( {2\alpha + \beta } \right)$ by substituting the values. And in this, we will solve this question.

Complete step by step solution:
So it is given that $\sin \left( {\alpha + \beta } \right) = 1$
And since, $\left( {\alpha ,\beta \in \left[ {0,\pi /2} \right]} \right)$ so from this
$ \Rightarrow \alpha + \beta = \pi /2$
Also, it is given that $\sin \left( {\alpha - \beta } \right) = 1/2$
And since, $\left( {\alpha ,\beta \in \left[ {0,\pi /2} \right]} \right)$ so from this
$ \Rightarrow \alpha - \beta = \pi /6$
From this, on solving for the value of $\alpha \& \beta $ , we get
$ \Rightarrow \alpha = \dfrac{\pi }{3}{\text{ and }}\beta = \dfrac{\pi }{6}$
Since we have to find $\tan \left( {\alpha + 2\beta } \right)\tan \left( {2\alpha + \beta } \right)$ so for this we will substitute the values of $\alpha \& \beta $ and we get
$ \Rightarrow \alpha + 2\beta = \dfrac{{2\pi }}{3}$
Therefore putting it in the main equation, we get
$ \Rightarrow \tan \left( {\alpha + 2\beta } \right) = \tan \dfrac{{2\pi }}{3}$
On further expanding more, the tangent will be
$ \Rightarrow \tan \left( {\pi - \pi /3} \right)$
And as we know that $\tan \left( {\pi - \theta } \right) = - \tan \theta $
Therefore, the above equation will be equal to
$ \Rightarrow - \tan \pi /3$
And on putting the values for it, we get
$ \Rightarrow - \sqrt 3 $
Therefore, we have $\tan \left( {\alpha + 2\beta } \right) = - \sqrt 3 $ , we will name it equation $1$
Also for this, we will substitute the values of $\alpha \& \beta $ and we get
$ \Rightarrow 2\alpha + \beta = \dfrac{{5\pi }}{6}$
Therefore putting it in the main equation, we get
$ \Rightarrow \tan \left( {2\alpha + \beta } \right) = \tan \dfrac{{5\pi }}{6}$
On further expanding more, the tangent will be
$ \Rightarrow \tan \left( {\pi - \pi /6} \right)$
And as we know that $\tan \left( {\pi - \theta } \right) = - \tan \theta $
Therefore, the above equation will be equal to
$ \Rightarrow - \tan \pi /6$
And on putting the values for it, we get
$ \Rightarrow - \dfrac{1}{{\sqrt 3 }}$
Therefore, we have $\tan \left( {2\alpha + \beta } \right) = - 1/\sqrt 3 $ , we will name it equation $2$
Now from equation $1$ and equation $2$ , on substituting the values for $\tan \left( {\alpha + 2\beta } \right)\tan \left( {2\alpha + \beta } \right)$ , we get
$ \Rightarrow - \sqrt 3 \times \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$
On canceling the like terms, we get
$ \Rightarrow 1$
Therefore, $\tan \left( {\alpha + 2\beta } \right)\tan \left( {2\alpha + \beta } \right)$ will be equal to $1$ .

Hence, the option $\left( a \right)$ is correct.

Note:
Here to make this simplification easier we can use the degree instead of radian as sometimes it may make someone stuck for a bit to think about it. Also, one more point we should take care of is we should remember the interval of the functions as it will always help to solve this type of problem.