Question

If $\sin \left(\theta \right)=3\sin \left(\theta +2\alpha \right)$, then the value of $\tan \left(\theta +\alpha \right)+2\tan \left(\alpha \right)$ is

• A. $3$
• B. $2$
• C. $-1$
• D. $0$
• E. $1$

Answer 1

The correct option is D

$0$

Explanation for the correct option.

Simplify the given equation like this,

$\sin \left(\theta +\alpha -a\right)=3\sin \left(\theta +\alpha +\alpha \right)$

Now apply the trigonometric identities,

Using $\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$ in the LHS and $\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$ in the RHS,

$\begin{array}{rcl}\therefore \sin \left(\theta +\alpha \right)\cos \left(\alpha \right)-\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)& =& 3\left[\sin \left(\theta +\alpha \right)\cos \left(\alpha \right)+\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)\right]\\ & \Rightarrow & \sin \left(\theta +\alpha \right)\cos \left(\alpha \right)-\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)=3\sin \left(\theta +\alpha \right)\cos \left(\alpha \right)+3\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)\\ & \Rightarrow & \sin \left(\theta +\alpha \right)\cos \left(\alpha \right)-3\sin \left(\theta +\alpha \right)\cos \left(\alpha \right)=3\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)+\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)\end{array}$

From this, we can see that,

$\begin{array}{rcl}\therefore -2\sin \left(\theta +\alpha \right)\cos \left(\alpha \right)& =& 4\sin \left(\alpha \right)\cos \left(\theta +\alpha \right)\\ & \Rightarrow & -\frac{\sin \left(\theta +\alpha \right)}{\cos \left(\theta +\alpha \right)}=2\frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}\\ & \Rightarrow & -\tan \left(\theta +\alpha \right)=2\tan \left(\alpha \right)\\ & \Rightarrow & \tan \left(\theta +\alpha \right)+2\tan \left(\alpha \right)=0\end{array}$

Hence, the correct option is (B).