We'll consider that x > 0. By the Pythagorean trigonometric identity, we have:
[tex]\sin^2\theta+\cos^2\theta=1\\\\
\left(\dfrac{x}{9}\right)^2+\cos^2\theta=1\\\\
\dfrac{x^2}{81}+\cos^2\theta=1\\\\
\cos^2\theta=1-\dfrac{x^2}{81}\\\\
\cos^2\theta=\dfrac{81-x^2}{81}\\\\
\cos\theta=\pm\sqrt{\dfrac{81-x^2}{81}}\\\\\cos\theta=\pm\dfrac{\sqrt{81-x^2}}{9}[/tex]
Since θ is in the 1st quadrant, [tex]\cos\theta\ \textgreater \ 0[/tex]. So:
[tex]\boxed{\cos\theta=\dfrac{\sqrt{81-x^2}}{9}}[/tex]
Now, we'll find the tangent. Using the formula below:
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\
\tan\theta=\dfrac{\dfrac{x}{9}}{~~\dfrac{\sqrt{81-x^2}}{9}~~}=\dfrac{x}{9}\times\dfrac{9}{\sqrt{81-x^2}}\\\\
\boxed{\tan\theta=\dfrac{x}{\sqrt{81-x^2}}}[/tex]
