Chapter 5.1, Problem 8E

To determine

To evaluate: the remaining trigonometric function of the angle.

Answer to Problem 8E

  $\sin x=-\frac{6}{7},\cos x=-\frac{\sqrt{13}}{7},\sec x=-\frac{7\sqrt{13}}{13},\tan x=\frac{6\sqrt{13}}{13},\cot x=\frac{\sqrt{13}}{6}$ .

Explanation of Solution

Given information:

Given trigonometric function is

  $\csc x=-\frac{7}{6},\tan x>0$ .

Calculation:

Find the trigonometric function $\sin x$ by applying reciprocal identity;

  $\csc x=\frac{1}{\sin x}$

Substitute $\csc x=-\frac{7}{6}$ in the above function;

  $\begin{array}{c}\csc x=-\frac{7}{6}\\ \frac{1}{\sin x}=-\frac{7}{6}\\ \sin x=-\frac{6}{7}\end{array}$

Now, apply Pythagorean Identity ${\cos }^{2}x=1-{\sin }^{2}x$ to find $\cos x$ .

  $\begin{array}{c}{\cos }^{2}x=1-{\sin }^{2}x\\ \cos x=\pm \sqrt{1-{\sin }^{2}x}\\ =\pm \sqrt{1-{\left(-\frac{6}{7}\right)}^2}\\ =\pm \sqrt{1-\frac{36}{49}}\\ =\pm \sqrt{\frac{13}{49}}\\ =\pm \frac{\sqrt{13}}{7}\end{array}$

As, $\sin x<0,\tan x>0$ so $\cos x<0$

Hence, $\cos x=-\frac{\sqrt{13}}{7}$

Apply reciprocal identity $\sec u=\frac{1}{\cos u}$

  $\sec x=\frac{1}{\cos x}$

Substitute $\cos x=-\frac{\sqrt{13}}{7}$ in the above expression;

  $\begin{array}{c}\sec x=\frac{1}{-\frac{\sqrt{13}}{7}}\\ =-\frac{7}{\sqrt{13}}\\ =-\frac{7}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ =-\frac{7\sqrt{13}}{13}\end{array}$

Hence, $\sec x=-\frac{7\sqrt{13}}{13}$ .

Now, use quotient identity $\tan u=\frac{\sin u}{\cos u}$ :

  $\begin{array}{c}\tan x=\frac{\sin x}{\cos x}\\ =\frac{-\frac{6}{7}}{-\frac{\sqrt{13}}{7}}\\ =\frac{6}{\sqrt{13}}\\ =\frac{6}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ =\frac{6\sqrt{13}}{13}\end{array}$

Hence, $\tan x=\frac{6\sqrt{13}}{13}$

Substitute $\tan x=\frac{6\sqrt{13}}{13}$ in the reciprocal identity $\cot x=\frac{1}{\tan x}$ ;

  $\begin{array}{c}\cot x=\frac{1}{\tan x}\\ =\frac{1}{\frac{6\sqrt{13}}{13}}\\ =\frac{13}{6\sqrt{13}}\\ =\frac{\sqrt{13}}{6}\end{array}$

Hence, $\cot x=\frac{\sqrt{13}}{6}$ .

Therefore remaining trigonometric functions are $\sin x=-\frac{6}{7},\cos x=-\frac{\sqrt{13}}{7},\sec x=-\frac{7\sqrt{13}}{13},\tan x=\frac{6\sqrt{13}}{13},\cot x=\frac{\sqrt{13}}{6}$ .