Question

The general solution of $3\tan (\theta -{15}^{\circ })=\tan (\theta +{15}^{\circ })$ is:

• A. $\theta =\frac{n\pi }{4}+(-1{)}^n\frac{\pi }{2};n\in Z$
• B. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4};n\in Z$
• C. $\theta =\frac{n\pi }{4}+(-1{)}^n\frac{\pi }{8};n\in Z$
• D. $\theta =\frac{n\pi }{4}+(-1{)}^n\frac{\pi }{4};n\in Z$
  

Answer 1

The correct option is B $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4};n\in Z$

The given equation is $3\tan (\theta -{15}^{\circ })=\tan (\theta +{15}^{\circ })$
$\Rightarrow \frac{\tan (\theta +{15}^{\circ })}{\tan (\theta -{15}^{\circ })}=\frac{3}{1}$
Using componendo and dividendo,
$\Rightarrow \frac{\tan (\theta +{15}^{\circ })+\tan (\theta -{15}^{\circ })}{\tan (\theta +{15}^{\circ })-\tan (\theta -{15}^{\circ })}=\frac{3+1}{3-1}=2$

$\Rightarrow \frac{\frac{\sin (\theta +{15}^{\circ })}{\cos (\theta +{15}^{\circ })}+\frac{\sin (\theta -{15}^{\circ })}{\cos (\theta -{15}^{\circ })}}{\frac{\sin (\theta +{15}^{\circ })}{\cos (\theta +{15}^{\circ })}-\frac{\sin (\theta -{15}^{\circ })}{\cos (\theta -{15}^{\circ })}}=2$

$\Rightarrow \frac{\sin (\theta +{15}^{\circ })\cos (\theta -{15}^{\circ })+\cos (\theta +{15}^{\circ })\sin (\theta -{15}^{\circ })}{\sin (\theta +{15}^{\circ })\cos (\theta -{15}^{\circ })-\cos (\theta +{15}^{\circ })\sin (\theta -{15}^{\circ })}=2$

Using $\sin A\cos B+\cos A\sin B=\sin (A+B)$
and
$\sin A\cos B-\cos A\sin B=\sin (A-B)$

$\Rightarrow \frac{\sin (\theta +{15}^{\circ }+\theta -{15}^{\circ })}{\sin (\theta +{15}^{\circ }-(\theta -{15}^{\circ }\left)\right)}=2$
$\Rightarrow \frac{\sin 2\theta }{\sin {30}^{\circ }}=2$
$\Rightarrow \sin 2\theta =1$
We know that
$\sin \theta =\sin \alpha \Rightarrow \theta =n\pi +(-1{)}^n\alpha ;n\in Z$

We also know that $\sin \frac{\pi }{2}=1$
So we can write,
$\Rightarrow 2\theta =n\pi +(-1{)}^n\frac{\pi }{2};n\in Z$

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