Question

Find the principal value of ${{\tan }^{-1}}\left( 1 \right)-{{\cos }^{-1}}\left( \dfrac{-1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$.

Answer 1

Hint:For the above question we have to find the principal value of each of the inverse trigonometric functions in the given expression. Principal value of an inverse trigonometric function at a point x is that value which lies in the principal range. As we know that principal range for ${{\sin }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$, for ${{\tan }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and for $co{{s}^{-1}}x$ is $\left[ 0,\pi \right]$. We can start solving by taking $\theta ={{\tan }^{-1}}(1)$ and then taking tan on both sides to find the value of $\theta$.

Complete step-by-step answer:
We have been given the expression ${{\tan }^{-1}}\left( 1 \right)-{{\cos }^{-1}}\left( \dfrac{-1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$.
Now let us suppose $\theta ={{\tan }^{-1}}(1)$
On taking tangent function to both sides, we get as follows:
$\tan \theta =\tan \left[ {{\tan }^{-1}}(1) \right]$
We know that $\tan {{\tan }^{-1}}x=x$
$\Rightarrow \tan \theta =1$
Since the principal range of ${{\tan }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$,
So ‘$\theta$’ must lie between $\dfrac{-\pi }{2}$ and $\dfrac{\pi }{2}$.
We know that $\tan \left( \dfrac{\pi }{4} \right)=1$
Also, $\dfrac{\pi }{4}\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$
$\Rightarrow {{\tan }^{-1}}(1)=\dfrac{\pi }{4}$
Thus let us suppose $\theta ={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)$
On taking cosine function to both sides on equation, we get as follows:
$\cos \theta =\cos \left[ {{\cos }^{-1}}\left( \dfrac{-1}{2} \right) \right]$
We know that $\cos \left( {{\cos }^{-1}}x \right)=x$.
$\Rightarrow \cos \theta =\dfrac{-1}{2}$
Since the principal range for ${{\cos }^{-1}}x$ is $\left[ 0,\pi \right]$,
So $\theta$ must lie between 0 to $\pi$ both included.
We know that $\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}$
Also, $\dfrac{2\pi }{3}\in \left[ 0,\pi \right]$
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{2\pi }{3}$
Again, let us suppose $\theta ={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$
On taking sine function to both sides, we get as follows:
$\sin \theta =\sin \left[ {{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \right]$
We know that $\sin \left[ {{\sin }^{-1}}x \right]=x$
$\Rightarrow \sin \theta =\dfrac{-1}{2}$
Since the principal range for ${{\sin }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.
So $\theta$ must lie between $\dfrac{-\pi }{2}$ and $\dfrac{\pi }{2}$ both values included.
We know that $\sin \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}$
$\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{-\pi }{6}$ as $\dfrac{-\pi }{6}\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
Now substituting the values of ${{\tan }^{-1}}(1),{{\cos }^{-1}}\left( \dfrac{-1}{2} \right),{{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$ in the given expression we get as follows:
${{\tan }^{-1}}\left( 1 \right)-{{\cos }^{-1}}\left( \dfrac{-1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}+\left( \dfrac{-\pi }{6} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}-\dfrac{\pi }{6}$
On taking LCM of the terms we get as follows:
${{\tan }^{-1}}\left( 1 \right)-{{\cos }^{-1}}\left( \dfrac{-1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{3\pi +8\pi -2\pi }{12}=\dfrac{9\pi }{12}=\dfrac{3\pi }{4}$
Therefore, the value of the given expression is equal to $\dfrac{3\pi }{4}$.

Note: Be careful while finding the values of inverse trigonometric functions and also check it once that the values must lie between the defined range of the corresponding function. Also remember if there are two values, one is positive and other is negative such that they are numerically same then the principal value of the trigonometric function will be positive one.