A) \[x-\frac{\pi }{6}\]
B) \[x+\frac{\pi }{6}\]
C) \[2x-\frac{\pi }{6}\]
D) \[2x-\frac{\pi }{3}\]
Correct Answer: A
Solution :
| \[2y={{\left( {{\cot }^{-1}}\left( \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right)}^{2}}\] |
| \[2y={{\left( {{\cot }^{-1}}\left( \frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x} \right) \right)}^{2}}\] |
| \[2y={{\left( {{\cot }^{-1}}\left( \tan \left( \frac{\pi }{3}+x \right) \right) \right)}^{2}}\] |
| \[2y={{\left( \frac{\pi }{2}-{{\tan }^{-1}}\left( \tan \left( \frac{\pi }{2}+x \right) \right) \right)}^{2}}\] |
| \[={{\left( \frac{\pi }{2}-\left( \frac{\pi }{3}+x \right) \right)}^{2}}\] |
| \[2y={{\left( x-\frac{\pi }{6} \right)}^{2}}\] |
| \[2y={{x}^{2}}-\frac{\pi }{3}+\frac{{{\pi }^{2}}}{36}\] |
| \[y'=x-\frac{\pi }{6}.\] |
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