MHT CET · Maths · Trigonometric Equations
If \(2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)\), then \(\tan \theta\)
- A \(\frac{-1}{\sqrt{3}}\)
- B \(-\sqrt{3}\)
- C \(\sqrt{3}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(-\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right) \)
\( 2\left[\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right]=\left(\cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6}\right) \)
\( \therefore 2\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=\left(\frac{\sqrt{3} \cos \theta}{2}+\frac{\sin \theta}{2}\right) \)
\( \therefore\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=0 \Rightarrow \sin \theta+\sqrt{3} \cos \theta=0 \Rightarrow\)\( \tan \theta=-\sqrt{3}\)
\( 2\left[\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right]=\left(\cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6}\right) \)
\( \therefore 2\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=\left(\frac{\sqrt{3} \cos \theta}{2}+\frac{\sin \theta}{2}\right) \)
\( \therefore\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=0 \Rightarrow \sin \theta+\sqrt{3} \cos \theta=0 \Rightarrow\)\( \tan \theta=-\sqrt{3}\)
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