Question

Given cos0=4/9 and csc0 <0find sin0 and tan0​

Answer 1

D

Explanation:

ok

Answer 2

Recall that

`\cos^2\theta+\sin^2\theta=1\implies\sin\theta=-√(1-\cos^2\theta)`

where we take the negative square root because we know
`\csc\theta=\frac1{\sin\theta}<0`. Then

`\sin\theta=-√(1-\left(\frac49\right)^2)=-\frac{√(65)}9`

Then by definition of tangent,

`\tan\theta=(\sin\theta)/(\cos\theta)=\frac{-\frac{√(65)}9}{\frac49}=-\frac{√(65)}4`