Question

I kinda forgot my trig (kinda)

Answer 1

$\sec^2\theta=1+\tan^2\theta\implies \sec\theta=\pm\sqrt{1+\tan^2\theta}$

Since $\cos\theta>0$ for $0$, we have $\sec\theta>0$, so we take the positive root. Now,

$\sec\theta=\sqrt{1+\left(\dfrac{12}5\right)^2}=\dfrac{13}5$

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

In the first quadrant, cosine is positive, so

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

and in turn,

$\sin\theta=\dfrac35\implies\cos\theta=\sqrt{1-\left(\dfrac35\right)^2}=\dfrac45$
$\implies\tan\theta=\dfrac{\frac35}{\frac45}=\dfrac34$

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In the previous problem, we had $\cos\theta=\dfrac45$, so we must have $\tan\theta=\dfrac34$, which means

$\cot\theta=\dfrac1{\tan\theta}=\dfrac1{\frac34}=\dfrac43$