$\tan \theta = \frac{\sqrt15}{3}$

$\cos \theta=\frac{\sqrt10}{4}$

We're supposed to solve for $\sin \theta$ using trig identities, so:













Using this method, $\sin{\theta}$ is somehow greater than one? The above answer simplifies to about $1.0206$.

1 + cot$^2\theta$ = csc$^2\theta$




$\frac{24}{15}$ = csc$^2\theta$





$\frac{2\sqrt10}{8}$ = sin $\theta$


This is the $\cos$ value (maybe I made a mistake?)

I have asked my math teacher about this. She is also very confused as to what the answer is: the worksheet she printed out wasn't hers, and the answer listed is the second one, which is greater than one and is probably wrong somehow. I realized I could use the third identity only after I met with her, but she couldn't find any problem with the first two methods, and yet somehow they yielded different values.

Could someone please shed some light as to why I ended up with multiple answers, with one greater than one? Thanks for any answers.


The problem has no solutions because there is no $\theta$ such that $\tan\theta=\frac{\sqrt{15}}3$ and that $\cos\theta=\frac{\sqrt{10}}4$. In fact, if $\cos\theta=\frac{\sqrt{10}}4$, then$$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\frac6{16}}{\frac{10}{16}}=\frac35,$$and therefore $\tan\theta=\sqrt{\frac35}$.


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