Question

If ${\displaystyle \sin (\theta +\frac{\pi }{6})=4\cos (\theta -\frac{\pi }{3})}$, then tan θ = ______

विकल्प

• ${\displaystyle \sqrt{3}}$

• ${\displaystyle -\frac{1}{\sqrt{3}}}$

• ${\displaystyle \frac{1}{\sqrt{3}}}$

• ${\displaystyle -\sqrt{3}}$

Answer 1

If ${\displaystyle \sin (\theta +\frac{\pi }{6})=4\cos (\theta -\frac{\pi }{3})}$, then tan θ = ${\displaystyle \underline{-\frac{1}{\sqrt{3}}}}$.

Explanation:

${\displaystyle \sin (\theta +\frac{\pi }{6})=4\cos (\theta -\frac{\pi }{3})}$

∴ ${\displaystyle \sin \theta .\cos \frac{\pi }{6}+\cos \theta .\sin \frac{\pi }{6}=4(\cos \theta .\cos \frac{\pi }{3}+\sin \theta .\sin \frac{\pi }{3})}$

⇒ ${\displaystyle \frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta =4(\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta )}$

⇒ ${\displaystyle \cos \theta +\sqrt{3}\sin \theta =0}$

⇒ tan θ = ${\displaystyle -\frac{1}{\sqrt{3}}}$