Chapter 5.1, Problem 13E

To determine

To calculate: The values of all six trigonometric functions.

Answer to Problem 13E

The required values of $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ are $\frac{1}{\sqrt{3}}$ , $\sqrt{3}$ , $\frac{2}{\sqrt{3}}$ and $2$ respectively.

Explanation of Solution

Given information:

The functions: $\sin x=\frac{1}{2}$ , $\cos x=\frac{\sqrt{3}}{2}$

Formula used:

Trigonometric identities:

  $\sin x=\frac{1}{\csc x}$

  $\sec x=\frac{1}{\cos x}$

  $\tan x=\frac{\sin x}{\cos x}$

  $\cot x=\frac{1}{\tan x}$

  ${\sin }^{2}x+{\cos }^{2}x=1$

  ${\sec }^{2}x=1+{\tan }^{2}x$

Calculation:

The given function is $\sin x=\frac{1}{2}$ , $\cos x=\frac{\sqrt{3}}{2}$ .

So, $\begin{array}{c}\csc x=\frac{1}{\sin x}\\ =\frac{1}{\frac{1}{2}}\\ =2\end{array}$

And, $\begin{array}{c}\sec x=\frac{1}{\cos x}\\ =\frac{1}{\frac{\sqrt{3}}{2}}\\ =\frac{2}{\sqrt{3}}\end{array}$

Use the formula $\tan x=\frac{\sin x}{\cos x}$ to find the value of $\tan x$ .

  $\begin{array}{c}\tan x=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ =\frac{1}{\cancel{2}}\times \frac{\cancel{2}}{\sqrt{3}}\\ =\frac{1}{\sqrt{3}}\end{array}$

So, $\begin{array}{c}\cot x=\frac{1}{\tan x}\\ =\frac{1}{\frac{1}{\sqrt{3}}}\\ =\sqrt{3}\end{array}$

Hence, the required values of $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ are $\frac{1}{\sqrt{3}}$ , $\sqrt{3}$ , $\frac{2}{\sqrt{3}}$ and $2$ respectively.