Chapter 7.3, Problem 70AYU

In Problems 59-82, solve each equation on the interval $0\le \theta <2\pi$.

$\tan \theta =\cot \theta$

To determine

To find: The trigonometric equation in the form of $\text{tan}\theta =\text{ cot}\theta ,0\le \theta <2\pi$.

Answer to Problem 70AYU

The solution set for the trigonometric equation $\text{tan}\theta =\text{ cot}\theta$.

$=\{\pi /4,3\pi /4\},0\le \theta <2\pi$

Explanation of Solution

Given:

Trigonometric equation $\text{tan}\theta =\text{ cot}\theta ,0\le \theta <2\pi$.

Calculation:

$\text{tan}\theta =\text{ cot}\theta$

By definition,

$=\text{ sin}\theta /\text{cos}\theta =\text{ cos}\theta /\text{sin}\theta$

Multiplying by $\text{sin}\theta$ and $\text{cos}\theta$ on each side,

${\text{sin}}^2\theta ={\text{ cos}}^2\theta$

Using the identity,

${\text{cos}}^2\theta =1-{\text{ sin}}^2\theta$

${\text{sin}}^2\theta =1-{\text{ sin}}^2\theta$

Adding ${\text{sin}}^2\theta$ on each side,

$2{\text{sin}}^2\theta =1$

Dividing by 2 on each side,

${\text{sin}}^2\theta =1/2$

$\text{sin}\theta =1/\surd 2>0$

Therefore, $\theta$ lies in I and II quadrants.

Angle $\theta$ for which sine value $=1/\surd 2=\pi /4,3\pi /4.0\le \theta <2\pi$.