[tex]\sec^2\theta=1+\tan^2\theta\implies \sec\theta=\pm\sqrt{1+\tan^2\theta}[/tex]
Since [tex]\cos\theta>0[/tex] for [tex]0<\theta<\dfrac\pi2[/tex], we have [tex]\sec\theta>0[/tex], so we take the positive root. Now,
[tex]\sec\theta=\sqrt{1+\left(\dfrac{12}5\right)^2}=\dfrac{13}5[/tex]
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[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}[/tex]
[tex]\cos^2\theta=1-\sin^2\theta[/tex]
In the first quadrant, cosine is positive, so
[tex]\cos\theta=\sqrt{1-\sin^2\theta}[/tex]
and in turn,
[tex]\sin\theta=\dfrac35\implies\cos\theta=\sqrt{1-\left(\dfrac35\right)^2}=\dfrac45[/tex]
[tex]\implies\tan\theta=\dfrac{\frac35}{\frac45}=\dfrac34[/tex]
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In the previous problem, we had [tex]\cos\theta=\dfrac45[/tex], so we must have [tex]\tan\theta=\dfrac34[/tex], which means
[tex]\cot\theta=\dfrac1{\tan\theta}=\dfrac1{\frac34}=\dfrac43[/tex]
