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.