Question

Prove that
(i) $\sin \left(\theta -\frac{\pi }{6}\right)+\cos \left(\theta -\frac{\pi }{3}\right)=\sqrt{3}.\sin \theta$
(ii) $If2\sin \left(\theta +\frac{\pi }{3}\right)=\cos \left(\theta -\frac{\pi }{6}\right),then\tan \theta +\sqrt{3}=0$

Answer 1

(i)$Sin(\theta -\pi /6)+cos(\theta -\frac{\pi }{3})=\sqrt{3}sin\theta$

LHS

$sin(\theta -\pi /6)+cos(\theta -\frac{\pi }{3})=sin\theta cos\pi /6-sin\pi /6cos\theta +cos\theta cos\frac{\pi }{3}+sin\theta sin\left(\frac{\pi }{3}\right)$

$=\frac{\sqrt{3}}{2}sin\theta -\frac{cos\theta }{2}+\frac{cos\theta }{2}+\frac{\sqrt{3}}{2}sin\theta$

$=\sqrt{3}sin\theta =RHS$

$\therefore$ LHS = LHS

$\Rightarrow sin(\theta -\frac{\pi }{6}+cos(\theta -\frac{\pi }{3}\left)\right)=\sqrt{3}sin\theta$

(ii) if

$2sin(\theta +\frac{\pi }{6})=cos(\theta -\frac{\pi }{6})$ then $tan\theta +\sqrt{3}=0$

$2[sin\theta cos\frac{\pi }{3}+cos\theta sin\frac{\pi }{3}]=cos\left(cos\frac{\pi }{6}\right)+sin\theta \left(sin\frac{\pi }{6}\right)$

$\Rightarrow 2\times sin\theta \times \frac{1}{2}+2cos\theta \times \frac{\sqrt{3}}{2}=cos\theta \times \frac{\sqrt{3}}{2}+sin\theta \times \frac{1}{2}$

$\Rightarrow sin\theta \times \frac{1}{2}+\frac{\sqrt{3}cos\theta }{2}=0$

$\Rightarrow \frac{sos\theta }{2}[\frac{sin\theta }{cos\theta }+\sqrt{3}]=0$

$\Rightarrow tan\theta +\sqrt{3}=0$