Question

If $\sin \theta =3\sin \left( \theta +2\alpha \right),$ then the value of $\tan \left( \theta +\alpha \right)+2\tan \alpha$ is:
A. 3
B. 2
C. 1
D. 0

Answer 1

Hint: Use the Componendo Dividendo rule in the given expression. Apply trigonometric identities and simplify the expression to get the expression as $\tan \left( \theta +\alpha \right)+2\tan \alpha$.

Complete step by step solution:
Given is the expression $\sin \theta =3\sin \left( \theta +2\alpha \right)$
$\therefore \dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=3.$
Let us use the Componendo Dividendo rule to solve the above expression.
Componendo Dividendo is a theorem on proportions which is used to perform calculations and reduce the number of steps.
According to Componendo Dividendo if $\dfrac{a}{b}=\dfrac{c}{d},$then it implies that $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}.......(1)$
Thus applying Componendo Dividendo rule in the expression in equation (1),
$\dfrac{\sin \theta }{\sin \left( \theta +2\alpha \right)}=\dfrac{3}{1}......(2)$
Where,

& a=\sin \theta \\\ & b=\sin \left( \theta +2\alpha \right) \\\ & c=3 \\\ & d=1 \\\ \end{aligned}$$ $$\begin{aligned} & \therefore \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} \\\ & \Rightarrow \dfrac{\sin \theta +\sin \left( \theta +2\alpha \right)}{\sin \theta -\sin \left( \theta +2\alpha \right)}=\dfrac{3+1}{3-1} \\\ & \Rightarrow \dfrac{\sin \theta +\sin \left( \theta +2\alpha \right)}{\sin \theta -\sin \left( \theta +2\alpha \right)}=\dfrac{4}{2}=2......(3) \\\ \end{aligned}$$ We know the trigonometric identities, $$\begin{aligned} & \operatorname{sinx}+siny=2sin\left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \\\ & \sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right) \\\ \end{aligned}$$ Let us apply these identities in equation (3). $$x=\theta $$and$$y=\left( \theta +2\alpha \right)$$. $$\therefore \dfrac{2\sin \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\cos \left( \dfrac{\theta -\theta -2\alpha }{2} \right)}{2\cos \left( \dfrac{\theta +\theta +2\alpha }{2} \right)\sin \left( \dfrac{\theta -\theta -2\alpha }{2} \right)}=2$$ By simplifying the expression, we get, $$\begin{aligned} & \Rightarrow \dfrac{\sin \left( \dfrac{2\theta +2\alpha }{2} \right)\cos \left( \dfrac{-2\alpha }{2} \right)}{\cos \left( \dfrac{2\theta +2\alpha }{2} \right)\sin \left( \dfrac{-2\alpha }{2} \right)}=2 \\\ & \Rightarrow \dfrac{\sin \left( \theta +\alpha \right)\cos \left( -\alpha \right)}{\cos \left( \theta +\alpha \right)\sin \left( -\alpha \right)}=2 \\\ \end{aligned}$$ The cosine is an even function, thus $$\cos (-\alpha )=\cos \alpha$$. The sine is an odd function, so $$sin(-\alpha )=-\sin \alpha $$. $$\dfrac{\sin \left( \theta +\alpha \right)\cos \left( \alpha \right)}{-\cos \left( \theta +\alpha \right)\sin \left( \alpha \right)}=2$$. By cross multiplying, we get, $$\Rightarrow \dfrac{\sin \left( \theta +\alpha \right)}{\cos \left( \theta +\alpha \right)}=\dfrac{-2\sin \alpha }{\cos \alpha }$$. We know that $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }.$$ $$\begin{aligned} & \therefore \tan (\theta +\alpha )=-2\tan \alpha \\\ & \Rightarrow \tan (\theta +\alpha )+2\tan \alpha =0 \\\ \end{aligned}$$ Thus we got the value of $$\tan (\theta +\alpha )+2\tan \alpha $$ as 0. Hence option D is the correct answer. Note: Remember the basic trigonometric identities like $$(sinA+sinB)$$ and $$(sinA-sinB)$$ which we have used here. They are very important for solving expressions like these. Just apply the formula and simplify it and you will get the answer.