Question

If $\sin \left(x+3a\right)=3\sin \left(a-x\right)$, then

• A. $\tan \left(x\right)=\tan \left(a\right)$
• B. $\tan \left(x\right)={\tan }^2\left(a\right)$
• C. $\tan \left(x\right)={\tan }^3\left(a\right)$
• D. $\tan \left(x\right)=3\tan \left(a\right)$

Answer 1

The correct option is C

$\tan \left(x\right)={\tan }^3\left(a\right)$

Explanation for the correct option

Here use the identity $\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$ in the Left-hand side (LHS )and identity $\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$ in the RHS.

$\begin{array}{rcl}\therefore \sin \left(x\right)\cos \left(3a\right)+\sin \left(3a\right)\cos \left(x\right)& =& 3\left(\sin \left(a\right)\cos \left(x\right)-\sin \left(x\right)\cos \left(a\right)\right)\\ & \Rightarrow & \sin \left(x\right)\cos \left(3a\right)+3\sin \left(x\right)\cos \left(a\right)=3\sin \left(a\right)\cos \left(x\right)-\sin \left(3a\right)\cos \left(x\right)\\ & \Rightarrow & \sin \left(x\right)\left(\cos \left(3a\right)+3\cos \left(a\right)\right)=\cos \left(x\right)\left(3\sin \left(a\right)-\sin \left(3a\right)\right)\end{array}$

We know that $\mathbf{cos}\mathbf{\left(}\mathbf{3}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{cos}}^{\mathbf{3}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$, we have

$\cos \left(3\mathrm{A}\right)+3\cos \left(\mathrm{A}\right)=4{\cos }^3\left(\mathrm{A}\right)$ and $\mathbf{sin}\mathbf{\left(}\mathbf{3}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{-}\mathbf{4}{\mathbf{sin}}^{\mathbf{3}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$, then

$4{\sin }^3\left(\mathrm{A}\right)=3\sin \left(\mathrm{A}\right)-\sin \left(3\mathrm{A}\right)$

Now, the equation $\begin{array}{rcl}\sin \left(x\right)\left(\cos \left(3a\right)+3\cos \left(a\right)\right)& =& \cos \left(x\right)\left(3\sin \left(a\right)-\sin \left(3a\right)\right)\end{array}$ can be written as

$\begin{array}{rcl}\therefore \sin \left(x\right)\left(4{\cos }^3\left(a\right)\right)& =& \cos \left(x\right)\left(4{\sin }^3\left(a\right)\right)\\ & \Rightarrow & \tan \left(x\right)={\tan }^3\left(a\right)\end{array}$

Hence, the correct option is (C).