If \tan^{-1}x-\cot ^{-1}\left ( \frac{1}{\sqrt{3}} \right ),x> 0, find the value of x and hence find the value of \sec ^{-1}\left ( \frac{2}{x} \right ).

 

 

 

 
 
 
 
 

Answers (1)

given:
         \tan^{-1}x-cot^{-1}x= \tan^{-1}\left ( \frac{1}{\sqrt{3}} \right ),x> 0
\Rightarrow \tan^{-1}x-\tan^{-1}\left ( \frac{1}{x} \right )= \tan^{-1}\left ( \frac{1}{\sqrt{3}} \right )\: \: \left [ \because \cot^{-1}x= \tan^{-1}\left ( \frac{1}{x} \right ),x> 0 \right ]
\Rightarrow \tan^{-1}\left ( \frac{x-\frac{1}{x}}{1+x\frac{1}{x}} \right )= \tan^{-1}\left ( \frac{1}{\sqrt{3}} \right )
\Rightarrow \frac{x^{2}-1}{2x}= \frac{1}{\sqrt{3}}
\Rightarrow \sqrt{3}x^{2}-2x-\sqrt{3}= 0\:\: \Rightarrow \sqrt{3}x^{2}-3x+x-\sqrt{3}= 0
\Rightarrow \sqrt{3}x\left ( x-\sqrt{3} \right )+1\left ( x-\sqrt{3} \right )= 0\: \: \Rightarrow \left ( x-\sqrt{3} \right )\left ( \sqrt{3}x+1 \right )= 0
\Rightarrow x=- \frac{1}{\sqrt{3}},\sqrt{3}\: \: \because\: x> 0,x= \sqrt{3}
\Rightarrow \sec^{-1}\left ( \frac{2}{x} \right )= \sec ^{-1}\left ( \frac{2}{\sqrt{3}} \right )\: \: \Rightarrow \: \sec ^{-1}\left ( \frac{2}{x} \right )= \sec ^{-1}\left ( \sec \frac{\pi }{6} \right )
\Rightarrow \sec ^{-1}\left ( \frac{2}{x} \right )= \frac{\pi }{6}

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