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Question: Find the value of \[2{\tan ^{ - 1}}x\] if \[x > 1,\]. A) \[{\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}...

Find the value of 2tan1x2{\tan ^{ - 1}}x if x>1,x > 1,.
A) cos1(1x21+x2){\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)
B) π+sin1(2x1+x2)\pi + {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)
C) π+tan1(2x1x2)\pi + {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)
D) πcos1(1x21+x2)\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)

Explanation

Solution

Hint: Substitute x=tanθx = \tan \theta and then see what value we are getting and after that substitute the same thing in each and every option to see which option correctly matches.

Complete step-by-step answer:
Substituting x=tanθx = \tan \theta in 2tan1x2{\tan ^{ - 1}}x
Now tan1(tanθ)=θ{\tan ^{ - 1}}(\tan \theta ) = \theta Therefore 2tan1(tanθ)=2θ2{\tan ^{ - 1}}(\tan \theta ) = 2\theta
So all we need to see that which one of these options can get me 2θ2\theta after substituting x=tanθx = \tan \theta
For option D:
πcos1(1x21+x2)\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)
After substitution it becomes
πcos1(1(tanθ)21+(tanθ)2)\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right)
From trigonometric identity we know that (1(tanθ)21+(tanθ)2)=cos2θ\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right) = \cos 2\theta
Therefore the whole thing become

= \pi - {\cos ^{ - 1}}\left( {\cos 2\theta } \right)\\\ = \pi - 2\theta \end{array}$$ Which is clearly the incorrect option. For Option C: After substitution it becomes $$\pi + {\tan ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 - {{(\tan \theta )}^2}}}} \right)$$ We know that $$\dfrac{{2\tan \theta }}{{1 - {{(\tan \theta )}^2}}} = \tan 2\theta $$ Therefore it becomes $$\begin{array}{l} = \pi + {\tan ^{ - 1}}\left( {\tan 2\theta } \right)\\\ = \pi + 2\theta \end{array}$$ Which is an incorrect option. For option B: After substitution it becomes $$\pi + {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{(\tan \theta )}^2}}}} \right)$$ We know that $$\left( {\dfrac{{2\tan \theta }}{{1 + {{(\tan \theta )}^2}}}} \right) = \sin 2\theta $$ Therefore the whole thing becomes $$\begin{array}{l} = \pi + {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\\\ = \pi + 2\theta \end{array}$$ Which again is incorrect For option A: After substitution it becomes $${\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right)$$ As in option D we already know that $${\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right) = 2\theta $$ Therefore it is clear that Option A is the correct option here. Note: We must check each and every option carefully as these types of questions may contain multiple correct options so make sure you check each one of them before writing the final answer.