Question

If we have an inverse trigonometric expression as ${{\tan }^{-1}}\left( \cot \theta \right)=2\theta$ then find the value of $\theta$.
A). $\pm \dfrac{\pi }{3}$
B). $\pm \dfrac{\pi }{4}$
C). $\pm \dfrac{\pi }{6}$
D). None of these.

Answer 1

We will write ${{\tan }^{-1}}\left( \cot \theta \right)=2\theta$ as $\cot \theta =\tan 2\theta$ and then we will use the tangent identity as follows $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and also we can write $\cot \theta =\dfrac{1}{\tan \theta }$ and then we will rearrange this accordingly to get the value of $\theta$.

Complete step-by-step solution
It is given in the question that ${{\tan }^{-1}}\left( \cot \theta \right)=2\theta$ and we have to find the value of $\theta$. Taking tangent on both sides of the given equation we get equation as $\tan \left( {{\tan }^{-1}}\left( \cot \theta \right) \right)=\tan 2\theta$, Now we know that ${{\operatorname{tantan}}^{-1}}\theta =\theta$ therefore we get the given equation modified as $\cot \theta =\tan 2\theta$.
Now we use the formula of $\tan 2\theta$ as $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and also use the conversion $\cot \theta =\dfrac{1}{\tan \theta }$ in the above modified equation, we get $\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\dfrac{1}{\tan \theta }$. Now cross-multiplying both sides, we get $2{{\tan }^{2}}\theta =1-{{\tan }^{2}}\theta$ on transposing similar terms together, that is, on the same side, we get $2{{\tan }^{2}}\theta +{{\tan }^{2}}\theta =1$, simplifying further, we get $3{{\tan }^{2}}\theta =1$ that is ${{\tan }^{2}}\theta =\dfrac{1}{3}$ or $\tan \theta =\pm \dfrac{1}{\sqrt{3}}$.
Now, we know that the value of $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$, thus replacing $\pm \dfrac{1}{\sqrt{3}}$ as $\pm \tan \dfrac{\pi }{6}$, we get $\tan \theta =\pm \tan \dfrac{\pi }{6}$, now cancelling the similar terms on both the sides of the equation, we get $\theta =\pm \dfrac{\pi }{6}$ as our answer.
Thus option c) is the correct answer.

Note: We can solve this same question in a different way as - ${{\tan }^{-1}}\left( \cot \theta \right)=2\theta$ therefore $\cot \theta =\tan 2\theta$. We know that cotangent can be written as $\tan \left( \dfrac{\pi }{2}-\theta \right)$ and $\tan \left( \dfrac{-\pi }{2}-\theta \right)$ because both have same value therefore we get $\tan 2\theta =\tan \left( \dfrac{\pi }{2}-\theta \right)=\tan \left( -\dfrac{\pi }{2}-\theta \right)$. On cancelling similar terms, we get $2\theta =\left( \dfrac{\pi }{2}-\theta \right)or\left( -\dfrac{\pi }{2}-\theta \right)$ transposing $-\theta$ from RHS to LHS, we get $3\theta =\pm \dfrac{\pi }{2}$ or $\theta =\pm \dfrac{\pi }{6}$.