Question

Prove that: $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$.

Answer 1

Hint: As we can see that the above question is related to trigonometry as tangent i.e. $\tan $ is a trigonometric ratio. We will use the trigonometric identities to solve this question. We know the formula of $\tan 2\theta $, i.e. $\tan 2\theta = \left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$. We will use the value of ${\tan ^{ - 1}}x$ and then with the help of the formula we will solve it.

Complete step-by-step answer:
Let us assume that ${\tan ^{ - 1}}x = \theta $.
So by putting this in the value we have $x = \tan \theta $.
Now we know the trigonometric identity: $\tan 2\theta = \left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$. In this identity we can transfer the tan from the left hand side of the right hand side: we can write it as $2\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$.
Now we can substitute the value $\tan \theta = x$ in the identity and it gives us $2\theta = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$.
Also we have assumed ${\tan ^{ - 1}}x = \theta $, so by putting this back in place of $\theta $, we can write $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$.
Hence it is proved that : $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$.

Note: Before solving this kind of question we should have the full knowledge of the trigonometric identities and their formulas. There is an alternate way to solve this question with the help of another formula i.e. ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \arctan \left( {\dfrac{{a + b}}{{1 - ab}}} \right)$. We can assume that the value of $a = b = x$, so by putting this in the formula we also get the answer.