Question

If $3\tan (\theta -{15}^o)=\tan (\theta +{15}^o),0^o<\theta <{90}^o$, then $\sin \theta +\cos \theta$ is equal to _________.

Answer 1

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

By componendo and dividendo

$\frac{tan(\theta +15°)+tan(\theta -15°)}{tan(\theta +15°)-tan(\theta -15°)}=\frac{3+1}{3-1}\Rightarrow \frac{\frac{sin(\theta -15°)}{cos(\theta +15°)}+\frac{sin(\theta -15°)}{cos(\theta -15°)}}{\frac{sin(\theta +15°)}{cos(\theta +15°)}-\frac{sin(\theta -15°)}{cos(\theta -15°)}}=2\Rightarrow \frac{sin(\theta +15°)cos(\theta -15°)+sin(\theta -15°)cos(\theta -15°)}{sin(\theta +15°)cos(\theta -15°)-cos(\theta +15°)sin(\theta -15°)}=2\Rightarrow \frac{sin(\theta +15°+\theta -15°)}{sin(\theta +15°-\theta +15°)}=2\Rightarrow \frac{sin2\theta }{sin30°}=2\Rightarrow sin2\theta =1\Rightarrow sin2\theta =sin\frac{pi}{2}\Rightarrow 2\theta =n\pi +(-1{)}^n\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}=45°sin45°+cos45°=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$