Question

For the principal values evaluate ${{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$.

Answer 1

Hint:We will take the values of $\tan \left( \dfrac{\pi }{4} \right)=1$ and $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ as a guide to solve the given question. Also, we have to apply the formulas which are used to find the angle as there is a need for this. The formulas that we are going to use are $\tan \left( x \right)=\tan \left( y \right)$ which results into $x=n\pi +y$ and $\cos \left( x \right)=\cos \left( y \right)$ which results into $x=2n\pi \pm y$. Here n is the number that belongs to integers.

Complete step-by-step answer:
First we will consider the expression ${{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)...(i)$ and focus on the term ${{\tan }^{-1}}\left( -1 \right)$. We will do substitution here and substitute ${{\tan }^{-1}}\left( -1 \right)$ equal to x. This can be written as ${{\tan }^{-1}}\left( -1 \right)=x$. Now, we will take the inverse tangent expression to the right side of the equal sign. This results in $\tan x=-1$. Now, at this step we will use the value of $\tan \left( \dfrac{\pi }{4} \right)=1$ and substitute it into $\tan x=-1$. Therefore, we have $\tan x=-\tan \left( \dfrac{\pi }{4} \right)$. The tangent trigonometric operation is negative in second and fourth quadrants only. Thus, in the second quadrant the equation $\tan x=-\tan \left( \dfrac{\pi }{4} \right)$ changes into $\tan x=\tan \left( \pi -\dfrac{\pi }{4} \right)$ or
$\begin{aligned} & \tan x=\tan \left( \dfrac{4\pi -\pi }{4} \right) \\\ & \Rightarrow \tan x=\tan \left( \dfrac{3\pi }{4} \right)...(ii) \\\ \end{aligned}$
And, in the fourth quadrant the equation $\tan x=-\tan \left( \dfrac{\pi }{4} \right)$ changes into $\tan x=\tan \left( -\dfrac{\pi }{4} \right)...(iii)$.
As we know the formula to find the angle x here is given by $\tan \left( x \right)=\tan \left( y \right)$ which results into $x=n\pi +y$. Now we will apply this formula to equation (ii) and (iii). Therefore, we will get equation (ii) as $x=\left( \dfrac{3\pi }{4} \right)$ and equation (iii) as $x=\left( -\dfrac{\pi }{4} \right)$. At this step we will consider the range of inverse tangent which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. Clearly, $x=\left( \dfrac{3\pi }{4} \right)$ does not belongs to range but $x=\left( -\dfrac{\pi }{4} \right)$ does. Therefore, the value of $x=\left( -\dfrac{\pi }{4} \right)$ is considered here. As ${{\tan }^{-1}}\left( -1 \right)=x$ thus, we have ${{\tan }^{-1}}\left( -1 \right)=\left( -\dfrac{\pi }{4} \right)$.
Now, we will focus on ${{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$ and we will substitute this value equal to y. Therefore, we will have ${{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=y$. Now, we will place the inverse cosine term to the right side of the expression. Thus, we get $-\dfrac{1}{\sqrt{2}}=\cos \left( y \right)$. As we know that the value of $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. Thus, we get $\cos \left( y \right)=-\cos \left( \dfrac{\pi }{4} \right)$. As cosine is negative only in the second and third quadrants so we will take it negative in the second quadrant. Since, the range of the inverse cosine is $\left[ 0,\pi \right]$. Thus, we get $\cos \left( y \right)=\cos \left( \pi -\dfrac{\pi }{4} \right)$ in the second quadrant. This results into $\cos \left( y \right)=\cos \left( \dfrac{4\pi -\pi }{4} \right)$ or, $\cos \left( y \right)=\cos \left( \dfrac{3\pi }{4} \right)$.
Now, we will use the formula which is given by $\cos \left( y \right)=\cos \left( \theta \right)$ which results in $y=2n\pi \pm \theta$. Therefore, we have $\cos \left( y \right)=\cos \left( \dfrac{3\pi }{4} \right)$ results into $y=2n\pi \pm \dfrac{3\pi }{4}$ or $y=\dfrac{3\pi }{4}$. Since, ${{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=y$ thus we have ${{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{3\pi }{4}$.
Hence, the required principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{3\pi }{4}$.
Now, we will substitute the values in the expression (i). Therefore, we have
$\begin{aligned} & {{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=-\dfrac{\pi }{4}+\dfrac{3\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{-\pi +3\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{2\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{2} \\\ \end{aligned}$
Hence, the required value of the expression ${{\tan }^{-1}}\left( -1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{2}$.

Note: Convert the inverse trigonometric terms into simple trigonometric terms while changing their places to either side of equal sign. In case of inverse cosine function we could have also used the fourth quadrant also instead of second quadrant. This is because cosine is negative in both these quadrants. But since the range of inverse cosine is $\left[ 0,\pi \right]$ so, we will also take the value which belongs to this interval. As $\dfrac{3\pi }{4}$ belongs to the interval of inverse cosine therefore, we have selected this value. One should be aware that we are talking about the range of inverse cosine instead of cosine.