Question

Solution of the equation $3\tan (\theta -15)=\tan (\theta +15)$ is

• A. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4}$
• B. $\theta =n\pi +(-1{)}^n\frac{\pi }{3}$
• C. $\theta =n\pi -\frac{\pi }{3}$
• D. $\theta =n\pi -\frac{\pi }{4}$

Answer 1

Answer: (A)

Given, $3\tan (\theta -15)=\tan (\theta +15).$
$\frac{\tan A}{\tan B}=\frac{3}{1}$,
where $A=\theta +15^{\circ },B=\theta -15^{\circ }$
On applying componendo and dividendo, we get
$\Rightarrow \frac{\tan A+\tan B}{\tan A-\tan B}=\frac{3+1}{3-1}$
$\Rightarrow \frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=2$
$\Rightarrow \frac{\sin (A+B)}{\sin (A-B)}=2$
$\Rightarrow \sin 2\theta =2\sin 30^{\circ }$
$\Rightarrow \sin 2\theta =2\cdot \frac{1}{2}=1=\sin \frac{\pi }{2}$
$\Rightarrow 2\theta =n\pi +(-1{)}^n\frac{\pi }{2}$
$\Rightarrow \theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4}$