Question

If \[\sin \left( \alpha +\beta \right)=1\] and \[\sin \left( \alpha -\beta \right)=\dfrac{1}{2}\], where \[0\le \alpha ,\beta \le \dfrac{\pi }{2}\], then find the values of\[\tan \left( \alpha +2\beta \right)\] and \[\tan \left( 2\alpha +\beta \right)\]

Answer 1

Hint: Given \[\sin \left( \alpha +\beta \right)=1\] and \[\sin \left( \alpha -\beta \right)=\dfrac{1}{2}\]. we have to find the values of \[\tan \left( \alpha +2\beta \right)\] and \[\tan \left( 2\alpha +\beta \right)\]. Given the values of sine we have to find the angles \[\alpha \],\[\beta \]. Then we have to submit in the tangent and get the solution.
\[\sin \left( \alpha +\beta \right)=1\]

Complete step-by-step answer:
We know that value of sine is 1 when \[\theta \] is \[{{90}^{\circ }}\] in the given interval \[0\le \alpha ,\beta \le \dfrac{\pi }{2}\].
\[\left( \alpha +\beta \right)={{90}^{\circ }}\]. . . . . . . . . . . . . . . . . . . . . . . . (1)
Similarly \[\sin \left( \alpha -\beta \right)=\dfrac{1}{2}\]
We know that value of sine is \[\dfrac{1}{2}\] when \[\theta \]is \[{{30}^{\circ }}\] in the given interval \[0\le \alpha ,\beta \le \dfrac{\pi }{2}\].
\[\left( \alpha -\beta \right)={{30}^{\circ }}\]. . . . . . . . . . . . . . . . . . . . . . . . . (2)
Adding both 1 and 2 we get,
\[2\alpha ={{120}^{\circ }}\]
\[\alpha ={{60}^{\circ }}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
\[\beta ={{30}^{\circ }}\]. . . . . . . . . . . . . . . . . . . . . (b)
Now substituting the values in the in \[\tan \left( \alpha +2\beta \right)\] and \[\tan \left( 2\alpha +\beta \right)\]
\[\tan \left( \alpha +2\beta \right)\] = \[\tan \left( {{60}^{\circ }}+{{60}^{\circ }} \right)\]
\[=\tan \left( {{120}^{\circ }} \right)\]
\[=-\sqrt{3}\]
\[=-1.732\]
Now substituting the values in \[\tan \left( 2\alpha +\beta \right)\]
\[\tan \left( 2\alpha +\beta \right)\] = \[\tan \left( {{120}^{\circ }}+{{30}^{\circ }} \right)\]
\[=\tan ({{150}^{\circ }})\]
\[=\dfrac{-1}{\sqrt{3}}\]
\[=-0.577\]

Note: We have to take care that sine can have value 1 in other intervals also so check the intervals properly and solve. We can get a doubt like in the answer it was not following the interval, the given interval holds good only for the question given.