Question

Evaluate the following
${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)+{{\tan }^{-1}}\left( -\sqrt{3} \right)+{{\tan }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)$

Answer 1

Hint:First of all, use $\sin \left( -\theta \right)=-\sin \theta$. Now use, ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$ in each term. Now find the angles at which $\tan \theta =\dfrac{1}{\sqrt{3}},\tan \theta =\sqrt{3}$ and $\tan \theta =1$ or the value of ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right),{{\tan }^{-1}}\left( \sqrt{3} \right)$ and ${{\tan }^{-1}}\left( 1 \right)$ and substitute these in the given expression to get the required answer.

Complete step-by-step answer:
In this question, we have to find the value of the expression ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)+{{\tan }^{-1}}\left( -\sqrt{3} \right)+{{\tan }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)$.
First of all, let us consider the expression given in the question,
$E={{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)+{{\tan }^{-1}}\left( -\sqrt{3} \right)+{{\tan }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)$
We know that $\sin \left( -\theta \right)=-\sin \theta$. By using this in the above expression, we get,
$E={{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)+{{\tan }^{-1}}\left( -\sqrt{3} \right)+{{\tan }^{-1}}\left( -\sin \dfrac{\pi }{2} \right)$
Also, we know that, ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$. By using this in the above expression, we get,
$E=-{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)+\left( -{{\tan }^{-1}}\sqrt{3} \right)+{{\tan }^{-1}}\left( -\sin \dfrac{\pi }{2} \right)$
$E=-{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\left( \sin \dfrac{\pi }{2} \right).....\left( i \right)$
Now, let us draw the table for trigonometric ratios of general angles.

From the above table, we get, $\sin \dfrac{\pi }{2}=1$. By using this in equation (i), we get,
$E=-{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\left( 1 \right).....\left( ii \right)$
Also from the above table, we can see that
$\tan \left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}}$
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=\dfrac{\pi }{6}$
$\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$
$\Rightarrow {{\tan }^{-1}}\left( \sqrt{3} \right)=\dfrac{\pi }{3}$
$\tan \left( \dfrac{\pi }{4} \right)=1$
$\Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$
So by substituting the value of ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right),{{\tan }^{-1}}\left( \sqrt{3} \right)$ and ${{\tan }^{-1}}\left( 1 \right)$ in equation (ii), we get, $E=-\dfrac{\pi }{6}-\dfrac{\pi }{3}-\dfrac{\pi }{4}$
By simplifying the above equation, we get,
$E=-\left( \dfrac{\pi }{6}+\dfrac{\pi }{3}+\dfrac{\pi }{4} \right)$
$E=\dfrac{-\left( 2\pi +4\pi +3\pi \right)}{12}$
$E=-\left( \dfrac{9\pi }{12} \right)$
$E=\dfrac{-3\pi }{4}$
Hence, we get the value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)+{{\tan }^{-1}}\left( -\sqrt{3} \right)+{{\tan }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)$ as $\dfrac{-3\pi }{4}$.

Note: In this question, some students make this mistake of taking ${{\tan }^{-1}}\left( -x \right)$ as $\pi -{{\tan }^{-1}}\left( x \right)$ like in case of ${{\cot }^{-1}}\left( -x \right)$ which is wrong because ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right)$. Also in the case of inverse trigonometric functions, students must take care that the angle they take must lie in the range of respective functions to get the correct answer. Also, students should memorize the table for general trigonometric ratios.