Question

Find the principal value of ${{\tan }^{-1}}\left( \sqrt{3} \right)-{{\sec }^{-1}}\left( -2 \right)$.

Answer 1

Hint:For the above question we have to know about the principal values of an inverse trigonometric function. Principal value of an inverse trigonometric function is a value that belongs to the principal branch of the range of the function. We know that the principal branch of range of ${{\sec }^{-1}}x$ is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\} and ${{\tan }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.

Complete step-by-step answer:
We have been given the expression ${{\tan }^{-1}}\left( \sqrt{3} \right)-{{\sec }^{-1}}\left( -2 \right)$.
Now we know that the principal value means the value which lies between the defined range of the function.
For ${{\sec }^{-1}}x$ the range is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
For ${{\tan }^{-1}}x$ the range is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.
We know that $\tan \dfrac{\pi }{3}=\sqrt{3}$
So by substituting the value of $\sqrt{3}$ in ${{\tan }^{-1}}\left( \sqrt{3} \right)$.
${{\tan }^{-1}}\left( \sqrt{3} \right)={{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)$
Since we know that ${{\tan }^{-1}}\left( \tan \theta \right)=\theta$,
where $\theta$ must lie between $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.
$\Rightarrow \tan \sqrt{3}=\dfrac{\pi }{3}$
We know that $\sec \dfrac{2\pi }{3}=-2$
So by substituting the value of (-2) in ${{\sec }^{-1}}\left( -2 \right)$, we get as follows:
${{\sec }^{-1}}\left( -2 \right)={{\sec }^{-1}}\left( \sec \dfrac{2\pi }{3} \right)$
Since we know the trigonometric properties that ${{\sec }^{-1}}\left( \sec \theta \right)=\theta$, where $\theta$ must lie between \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
$\Rightarrow {{\sec }^{-1}}\left( -2 \right)=\dfrac{2\pi }{3}$
Now substituting the principal values of $\tan \sqrt{3}=\dfrac{\pi }{3}$ and ${{\sec }^{-1}}\left( -2 \right)=\dfrac{2\pi }{3}$ in the given expression, we get as follows:
$\tan \sqrt{3}-{{\sec }^{-1}}\left( -2 \right)=\dfrac{\pi }{3}-\dfrac{2\pi }{3}$
On taking LCM of terms, we get as follows:
$\tan \sqrt{3}-{{\sec }^{-1}}\left( -2 \right)=\dfrac{\pi -2\pi }{3}=\dfrac{-\pi }{3}$
Therefore, the principal value of the given expression $\tan \sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$ is equal to $\dfrac{-\pi }{3}$.

Note: Be careful while finding the principal value of inverse trigonometric functions and do check once that the value must lie between the principal branch of range of the function. Also, don’t confuse while finding the principal value of ${{\sec }^{-1}}\left( -2 \right)$ as sometimes by mistake we write ${{\sec }^{-1}}\left( -2 \right)=\dfrac{-\pi }{3}$ which is wrong because $\dfrac{-\pi }{3}$ doesn’t lie in the range of ${{\sec }^{-1}}x$ i.e. \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\} So be careful while finding it.