[tex]\cot\theta=7[/tex], so you immediately get
[tex]\dfrac1{\cot\theta}=\boxed{\tan\theta=\dfrac17}[/tex]
Recall the Pythagorean identity:
[tex]\tan^2\theta+1=\sec^2\theta\implies\sec\theta=\pm\sqrt{1-\left(\dfrac17\right)^2}=\pm\dfrac{4\sqrt3}7[/tex]
and from this we also get cosine for free, since
[tex]\dfrac1{\sec\theta}=\cos\theta=\pm\dfrac7{4\sqrt3}[/tex]
But only one of these can be correct. By definition of tangent,
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}[/tex]
For [tex]\theta[/tex] between π and 2π, we expect [tex]\sin\theta[/tex] to be negative. We konw [tex]\tan\theta[/tex] is positive, which means [tex]\cos\theta[/tex] must also be negative. So we have
[tex]\boxed{\cos\theta=\dfrac7{4\sqrt3}}[/tex]
[tex]\boxed{\sec\theta=\dfrac{4\sqrt3}7}[/tex]
and we can find sine using the tangent:
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}\implies\dfrac{\frac7{4\sqrt3}}7=\boxed{\sin\theta=\dfrac1{4\sqrt3}}[/tex]
and for free we get
[tex]\dfrac1{\sin\theta}=\boxed{\csc\theta=4\sqrt3}[/tex]
