JEE Main On 8 April 2017 Question 2
Question: If $ S={ x\in [0,2\pi ]: \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \\ \end{vmatrix} =0 }, $ Then $ \sum\limits _{x\in S}^{{}}{\tan }( \frac{\pi }{3}+x ) $ [JEE Online 08-04-2017]
Options:
A) $ 4+2\sqrt{3} $
B) $ -2-\sqrt{3} $
C) $ -2+\sqrt{3} $
D) $ -4-2\sqrt{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 0(0-cosx)-cosx(0-cos^{2}x)-sinx $ $ (sin^{2}x-0)=0 $ $ {{\cos }^{3}}x-{{\sin }^{3}}x=0 $ $ {{\tan }^{3}}=1\Rightarrow \tan x=1 $ $ \sum\limits _{{}}^{{}}{\frac{\sqrt{3}+\tan x}{1-\sqrt{3}}} $ $ \sum\limits _{{}}^{{}}{\frac{\sqrt{3}+1}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}}\Rightarrow \sum\limits _{{}}^{{}}{\frac{1+3+2\sqrt{3}}{-2}=}\sum\limits _{{}}^{{}}{\frac{4^{2}}{-2}-\frac{2\sqrt{3}}{3}} $ $ \sum\limits _{{}}^{{}}{-2-\sqrt{3}} $
