Question

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

Answer 1

Recall that

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

where we take the negative square root because we know $\csc\theta=\dfrac1{\sin\theta}$. Then

$\sin\theta=-\sqrt{1-\left(\dfrac49\right)^2}=-\dfrac{\sqrt{65}}9$

Then by definition of tangent,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt{65}}9}{\frac49}=-\dfrac{\sqrt{65}}4$