Question

Please help ASAP!

Answer 1

First, recall that

$\sin(-\theta)=-\sin\theta$

So if $\sin(-\theta)=\dfrac15$, then $\sin\theta=-\dfrac15$.

Second,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

We know that $\tan\theta>0$ and $\sin\theta$, which means we should also have $\cos\theta$.

Third,

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

but as we've already shown, we need to have $\cos\theta$, so we pick the negative root.

Finally,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\iff\dfrac6{\sqrt{12}}=\dfrac{-\frac15}{\cos\theta}\implies\cos\theta=-\dfrac1{5\sqrt3}$

Unfortunately, none of the given answers match, so perhaps I've misunderstood one of the given conditions... In any case, this answer should tell you everything you need to know to find the right solution from the given options.