Question

If $3\tan (\theta -{15}^{\circ })=\tan (\theta +{15}^{\circ })$, then $\theta$ is equal to $\left(nϵz\right)$

• A. $n\pi +\frac{\pi }{4}$
• B. $n\pi +\frac{\pi }{8}$
• C. $n\pi +\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is A $n\pi +\frac{\pi }{4}$

As $\tan 15=\tan (45-30)=\frac{\tan 45-\tan 30}{1+\tan 45\tan 30}$

$=\frac{1-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Therefore

$3\tan (\theta -15)=\tan (\theta +15)$

$\Rightarrow 3\frac{\tan \theta -\tan 15}{1+\tan \theta \tan 15}=\frac{\tan \theta +\tan 15}{1-\tan \theta \tan 15}$

$\Rightarrow (3\tan \theta -3\tan 15)(1-\tan \theta \tan 15)=(\tan \theta +\tan 15)(1+\tan \theta \tan 15)$

$\Rightarrow 3\tan \theta -3{\tan }^2\theta \tan 15-3\tan 15+3\tan \theta {\tan }^{2}15=\tan \theta +{\tan }^2\theta \tan 15+\tan 15+\tan \theta {\tan }^{2}15$

$\Rightarrow 2\tan \theta -4{\tan }^2\theta \tan 15-4\tan 15+2\tan \theta {\tan }^{2}15=0$

$\Rightarrow 2\tan \theta -4{\tan }^2\theta (2-\sqrt{3})-4(2-\sqrt{3})+2\tan \theta (2-\sqrt{3}{)}^2=0$

$\Rightarrow 2\tan \theta -{\tan }^2\theta (8-4\sqrt{3})-(8-4\sqrt{3})+2\tan \theta (7-4\sqrt{3})=0$

$\Rightarrow {\tan }^2\theta (8-4\sqrt{3})-2\tan \theta (8-4\sqrt{3})+(8-4\sqrt{3})=0$

$\Rightarrow {\tan }^2\theta -2\tan \theta +1=0\Rightarrow \tan \theta =\frac{2\pm \sqrt{4-4}}{2}=1\Rightarrow \theta =n\pi +\frac{\pi }{4}$