Answer:
[tex]\tan\theta=\sqrt{15}[/tex]
Step-by-step explanation:
In quadrant I, all the trigonometric ratios are positive.
It was given that;
[tex]\cos \theta=\frac{1}{4}[/tex]
[tex]\implies \sec\theta=4[/tex]
We use the Pythagorean identity;
[tex]\sec^2\theta=1+tan^2\theta[/tex]
[tex]4^2=1+tan^2\theta[/tex]
[tex]16-1=tan^2\theta[/tex]
[tex]15=tan^2\theta[/tex]
[tex]\tan\theta=\pm\sqrt{15}[/tex]
since we are in the first quadrant; [tex]\tan\theta=\sqrt{15}[/tex]
