Question

Write the principal value of $\left[ {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \right]$

Answer 1

Hint: To solve this type of problems we have to know the inverse trigonometric functions like ${{\cos }^{-1}}(\cos \theta )$ and ${{\tan }^{-1}}(\tan \theta )$. By applying formulas and writing the values of trigonometric functions we will get the values in the range of a function.

Complete step-by-step answer:

Now writing the expression
$\left[ {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \right]$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that
$\cos {{60}^{\circ }}=\dfrac{1}{2}=\cos \dfrac{\pi }{3}$
$\tan {{45}^{\circ }}=1=\dfrac{\pi }{4}$
${{\cos }^{-1}}(-\theta )=\pi -{{\cos }^{-1}}\theta$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Now rewriting the expression (1) we get
$\left[ {{\tan }^{-1}}(\tan \dfrac{\pi }{4})+\pi -{{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right) \right]$
We know that
${{\cos }^{-1}}(\cos \theta )=\theta \ for all \theta \in \left[ 0,\pi \right]$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
${{\tan }^{-1}}(\tan \theta )=\theta \ for all \theta \in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ . . . . . . . . . . . . . . . . . . (c)
Now from (b) and (c) further writing the expression we get,
= $\left[ \dfrac{\pi }{4}+\pi -\dfrac{\pi }{3} \right]$
= $\left[ \dfrac{11\pi }{12} \right]$

Note: The interval $\left[ 0,\pi \right]$as shown in (b) is the range for the inverse trigonometric function ${{\cos }^{-1}}\theta$. From (a) the range of ${{\cos }^{-1}}\theta$is $\left[ 0,\pi \right]$we have to use that expression of (a) because the range of ${{\cos }^{-1}}\theta$is $\left[ 0,\pi \right]$. The range of ${{\tan }^{-1}}\theta$is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. Take care while doing calculations.