Question

If $3\tan (\theta -{15}^o)=\tan (\theta +{15}^o),0<\theta <{90}^o$, then $\theta ={45}^o$

• A. True
• B. False

Answer 1

The correct option is A True

$3\tan (\theta -{15}^{\circ })=\tan (\theta +{15}^{\circ })$

$\Rightarrow \frac{3}{1}=\frac{\tan (\theta +{15}^{\circ })}{\tan (\theta -{15}^{\circ })}$

Using componendo and dividendo rule, we have

$\frac{3+1}{3-1}=\frac{\tan (\theta +{15}^{\circ })+\tan (\theta -{15}^{\circ })}{\tan (\theta -{15}^{\circ })-\tan (\theta -{15}^{\circ })}$

$\Rightarrow \frac{4}{2}=\frac{\tan (\theta +{15}^{\circ })+\tan (\theta -{15}^{\circ })}{\tan (\theta +{15}^{\circ })-\tan (\theta -{15}^{\circ })}$

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

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

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

$\Rightarrow \frac{\sin 2\theta }{\sin {30}^{\circ }}=2$

$\Rightarrow \sin 2\theta =2\sin {30}^{\circ }$

$\Rightarrow \sin 2\theta =2\times \frac{1}{2}$

$\Rightarrow \sin 2\theta =1$

$\Rightarrow \sin 2\theta =\sin {90}^{\circ }$

$\Rightarrow 2\theta ={90}^{\circ }$

$\Rightarrow \theta =\frac{{90}^{\circ }}{2}$

$\therefore \theta ={45}^{\circ }$

Hence the given statement is true.