Question

Can somebody please help me? (PreCalc)

Answer 1

`\csc\theta=\frac1{\sin\theta}=\frac{√(17)}}4\implies\sin\theta=\frac4{√(17)}`

Recall that
`\sin^2\theta+\cos^2\theta=1`, which means


`\cos\theta=\pm√(1-\sin^2\theta)=\pm\sqrt{1-(16)/(17)}=\pm\frac1{√(17)}`

Since
`\tan\theta=(\sin\theta)/(\cos\theta)>0`, and
`\sin\theta>0`, it follows that
`\cos\theta>0` too, so you take the positive root:


`\cos\theta=\frac1{√(17)}`

Therefore


`\cot\theta=(\cos\theta)/(\sin\theta)=\frac{\frac1{√(17)}}{\frac4{√(17)}}=\frac14`