Question

If an angle $\alpha$ is divided into two parts $A\mathrm{and}B$ such that $A–B=x$ and $\tan A:\tan B=2:1$, then what is $\sin x$ equal:

• A. $3\sin \alpha$
• B. $\frac{2\sin \alpha }{3}$
• C. $\frac{\sin \alpha }{3}$
• D. $2\sin \alpha$

Answer 1

The correct option is C

$\frac{\sin \alpha }{3}$

Explanation for the correct option:

Solving the given equations.

$$\begin{gathered} A+B=\alpha ...\left(1\right) \\ A-B=x...\left(2\right) \end{gathered}$$

From $\left(1\right)\mathrm{and}\left(2\right)$

$$\begin{gathered} \begin{array}{rcl}2A& =& \alpha +x\\ & \Rightarrow & A=\frac{\alpha +x}{2}\end{array} \\ \Rightarrow B=\begin{array}{rcl}& & \frac{\alpha -x}{2}\end{array} \end{gathered}$$

Solving $\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{A}}{\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{B}}$:

$\begin{array}{rcl}\frac{\tan A}{\tan B}& =& \frac{\sin A}{\cos A}\times \frac{\cos B}{\sin B}\\ & =& \frac{\sin A·\cos B}{\cos A·\sin B}\\ & =& \frac{\sin \left({\displaystyle \frac{\alpha +x}{2}}\right)·\cos \left({\displaystyle \frac{\alpha -x}{2}}\right)}{\cos \left({\displaystyle \frac{\alpha +x}{2}}\right)·\sin \left({\displaystyle \frac{\alpha -x}{2}}\right)}\end{array}$

By multiplying both numerator and denominator by $2$, we get

$$\begin{gathered} \begin{array}{rcl}& & \frac{2\sin \left({\displaystyle \frac{\alpha +x}{2}}\right)·\cos \left({\displaystyle \frac{\alpha -x}{2}}\right)}{2\cos \left({\displaystyle \frac{\alpha +x}{2}}\right)·\sin \left({\displaystyle \frac{\alpha -x}{2}}\right)}\\ & =& \frac{\sin \alpha +\sin x}{\sin \alpha -\sin x}\left[\mathrm{By},\sin A+\sin B=2\sin (A+B)/2·\cos (A-B)/2 \\ \sin A-\sin B=2\cos (A+B)/2.\sin (A-B)/2\right]\end{array} \end{gathered}$$

Solving $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathbf{:}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{1}$:

$$\begin{gathered} \tan A:\tan B=2:1 \\ \Rightarrow \frac{\tan A}{\tan B}=\frac{2}{1} \\ \Rightarrow \frac{\sin \alpha +\sin x}{\sin \alpha -\sin x}=\frac{2}{1} \\ \Rightarrow \sin \alpha +\sin x=2\left(\sin \alpha -\sin x\right) \\ \Rightarrow \sin \alpha +\sin x=2\sin \alpha -2\sin x \\ \Rightarrow 3\sin x=\sin \alpha \\ \therefore \sin x=\frac{\sin \alpha }{3} \end{gathered}$$

Hence, the correct answer is option C.