Question

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

Answer 1

Hint: We will find the value of $ {{\tan }^{-1}}\left( 1 \right) $ . We will check if this value lies in the range of the function $ {{\tan }^{-1}}x $ . After that we will find the value of $ {{\cos }^{-1}}\left( -\dfrac{1}{2} \right) $ . We will check if this value lies in the range of the inverse cosine function. If both these values lie in the respective range, then they are the principal values. We will add these principal values to find the principal value of the given function.

Complete step by step answer:
The given function is $ {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right) $ . Let us look at the first term of the given function. The first term is $ {{\tan }^{-1}}\left( 1 \right) $ . We know that $ \tan \dfrac{\pi }{4}=1 $ . Therefore, we have $ {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4} $ . The range of the inverse tangent function is $ \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) $ . Since $ \dfrac{\pi }{4}\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) $ , it is the principal value of $ {{\tan }^{-1}}\left( 1 \right) $ .
Next, let us look at the second term of the given function. The second term is $ {{\cos }^{-1}}\left( -\dfrac{1}{2} \right) $ . We know that $ \cos \dfrac{2\pi }{3}=-\dfrac{1}{2} $ . Hence, we have $ {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3} $ . Now, we are aware that the range of the inverse cosine function is $ \left[ 0,\pi \right] $ . We can see that $ \dfrac{2\pi }{3}\in \left[ 0,\pi \right] $ . Therefore, $ {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3} $ is its principal value.
So, the principal value of the given function is the following,
 $ \begin{align}
  & {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3} \\
 & \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi +8\pi }{12} \\
 & \therefore {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{11\pi }{12} \\
\end{align} $
Hence, the principal value of the given function is $ \dfrac{11\pi }{12} $ .

Note:
 We should be aware of the range of inverse trigonometric functions. The inverse value of a trigonometric function can have multiple values. Therefore, it is important to define the principal value. The principal value makes the inverse trigonometric function single valued. This is essential for questions of this type.