Answer:
[tex]\sqrt{15}[/tex]
Step-by-step explanation:
[tex]\cos(\theta)=\frac{1}{4}[/tex] is given.
My attempt will to be to use the Pythagorean Identity:
[tex]\cos^2(\theta)+\sin^2(\theta)=1[/tex]
[tex](\frac{1}{4})^2+\sin^2(\theta)=1[/tex]
[tex]\frac{1}{16}+\sin^2(\theta)=1[/tex]
Subtract 1/16 on both sides:
[tex]\sin^2(\theta)=\frac{15}{16}[/tex]
We have the following quotient identity:
[tex]\frac{\sin^2(\theta)}{\cos^2(\theta)}=\tan^2(\theta)[/tex]
[tex]\frac{\sin^2(\theta)}{\cos^2(\theta)}=\frac{\frac{15}{16}}{\frac{1}{16}}[/tex]
Multiply the numerator and denominator by 16. This is multiplying the fraction by 1 so this doesn't change the value of the fraction.
[tex]\tan^2(\theta)=\frac{15}{1}[/tex]
[tex]\tan^2(\theta)=15[/tex]
Square root both sides:
[tex]\tan(\theta)=\pm \sqrt{15}[/tex]
Since we are in the first quadrant then all 6 trigonometric functions are positive there so the answer is [tex]\sqrt{15}[/tex].
