Question

$2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$, then $\tan \theta=$

• A. $-\frac{1}{\sqrt{3}}$
• B. $-\sqrt{3}$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Hint:

Alternatively, use the co-function identity: $\cos\left(\theta - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2} - (\theta - \frac{\pi}{6})\right) = \sin\left(\frac{2\pi}{3} - \theta\right)$. You can then expand this to avoid dealing with both sines and cosines initially!

Answer 1

The Correct Option is B

Step 1: Read the equation carefully.
We are told that $2\sin\left(\theta+\frac{\pi}{3}\right)=\cos\left(\theta-\frac{\pi}{6}\right)$, and we want $\tan\theta$. The neat trick here is to notice the two angles are linked.
Step 2: Spot the hidden link between the angles.
Look at $\theta+\frac{\pi}{3}$ and $\theta-\frac{\pi}{6}$. Their difference is exactly $\frac{\pi}{2}$, a right angle. So one angle is $90^\circ$ more than the other.
Step 3: Use the co-function rule.
Since $\left(\theta+\frac{\pi}{3}\right)=\left(\theta-\frac{\pi}{6}\right)+\frac{\pi}{2}$, and $\sin\left(A+\frac{\pi}{2}\right)=\cos A$, we get $\sin\left(\theta+\frac{\pi}{3}\right)=\cos\left(\theta-\frac{\pi}{6}\right)$.
Step 4: Substitute this back in.
The given equation becomes $2\cos\left(\theta-\frac{\pi}{6}\right)=\cos\left(\theta-\frac{\pi}{6}\right)$, which forces $\cos\left(\theta-\frac{\pi}{6}\right)=0$.
Step 5: Solve that simple condition.
If $\cos\left(\theta-\frac{\pi}{6}\right)=0$, then $\theta-\frac{\pi}{6}=\frac{\pi}{2}$, so $\theta=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3}$.
Step 6: Find the tangent.
Now $\tan\theta=\tan\frac{2\pi}{3}=\tan(180^\circ-60^\circ)=-\tan 60^\circ=-\sqrt{3}$. So the answer is option (2). \[ \boxed{\tan\theta=-\sqrt{3}} \]