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} $
$\Rightarrow \theta=\frac{ n \pi}{2}+(-1)^{ n } \frac{\pi}{4}$