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Mathematics Question on Trigonometric Identities

If 2sin⁡(θ+π3)=cos⁡(θ−π6),2 \sin \left( \theta + \frac{\pi}{3}\right) = \cos \left( \theta -\frac{\pi}{6}\right) , then tan⁡ θ=\tan \, \theta =

A

3\sqrt{3}

B

−13- \frac{1}{\sqrt{3}}

C

13\frac{1}{\sqrt{3}}

D

−3- \sqrt{3}

Answer

−3- \sqrt{3}

Explanation

Solution

We have, 2sin⁡(θ+π3)=cos⁡(θ−π6)2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)
⇒2[sin⁡θcos⁡π3+cos⁡θsin⁡π3]\Rightarrow 2\left[\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right]
=cos⁡θcos⁡π6+sin⁡θsin⁡π6= \cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6}
⇒2[sin⁡θ2+3cos⁡θ2]=32cos⁡θ+12sin⁡θ\Rightarrow 2\left[\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right]=\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta
⇒2sin⁡θ+23cos⁡θ=3cos⁡θ+sin⁡θ\Rightarrow 2 \sin \theta+2 \sqrt{3} \cos \theta=\sqrt{3} \cos \theta+\sin \theta
⇒sin⁡θ=−3cos⁡θ\Rightarrow \sin \theta=-\sqrt{3} \cos \theta
⇒tan⁡θ=−3\Rightarrow \tan \theta=-\sqrt{3}