Chapter 7.3, Problem 98AYU

To determine

To solve: To find the values of trigonometric expression $f\left(x\right)=\cot x$ where $f\left(x\right)=-\sqrt{3}$ .

Answer to Problem 98AYU

The values of trigonometric expression where $f\left(x\right)=\cot x$ since $f\left(x\right)=-\sqrt{3}$ is $n\pi +\frac{5\pi }{6}$ where n is any integer.

Explanation of Solution

Given information:

The mentioned trigonometric expression is $f\left(x\right)=\cot x$ where $f\left(x\right)=-\sqrt{3}$ .

Formula used:

For trigonometric formula $\tan \theta =\tan \alpha \Rightarrow \theta =n\pi -\alpha$ .

Also the trigonometric formula $\tan \theta =x\Rightarrow \theta ={\tan }^{-1}x$ and $\frac{1}{\cot x}=\tan x$ .

Calculation:

Consider the mentioned trigonometric expression $f\left(x\right)=\cot x$ .

Recall that for $f\left(x\right)=-\sqrt{3}$ the trigonometric formula $f\left(x\right)=\cot x$ is equal to $\cot x=-\sqrt{3}$ .

Therefore we equate the mentioned trigonometric expression $f\left(x\right)=\cot x$ equal to $f\left(x\right)=-\sqrt{3}$ .

Therefore $\cot x=-\sqrt{3}\Rightarrow x=-{\cot }^{-1}\sqrt{3}$ .

Recall that the trigonometric formula $\frac{1}{\cot x}=\tan x$ .

Therefore $\frac{1}{\cot x}=-\frac{1}{\sqrt{3}}\Rightarrow \tan x=-\frac{1}{\sqrt{3}}$

Recall that the trigonometric formula $\tan \theta =\tan \alpha \Rightarrow \theta =n\pi -\alpha$

Therefore $\tan x=-\frac{1}{\sqrt{3}}\Rightarrow \tan x=\tan \left(-\frac{\pi }{6}\right)\Rightarrow \tan x=\tan \left(\pi -\frac{\pi }{6}\right)\Rightarrow \tan x=\tan \frac{5\pi }{6}\Rightarrow x=n\pi +\frac{5\pi }{6}$ .

Therefore values of trigonometric expression $f\left(x\right)=\cot x$ Where $f\left(x\right)=-\sqrt{3}$ is $n\pi +\frac{5\pi }{6}$ where n is any integer.