<s>[INST] find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ 
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$

The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+\frac{81}{4}}}=\frac{2}{\sqrt{85}}
$$
$$
\cos\theta=\frac{\sin\theta}{\tan\theta}=\frac{\frac{3}{4}}{\frac{9}{2}}=\frac{3}{4}.\frac{2}{9}=\frac{1}{6}
$$
Why is it getting confused here ?
 [/INST] The problem is actually incorrect. Here is why: 
Consider a right triangle and angle $\theta$ in that triangle such that $\sin\theta=\dfrac{3}{4}.$
That means that the right triangle has one leg $3$ and hypotenuse $4$. This implies that the other leg is $\sqrt{4^2-3^2}=\sqrt{7}.$
This implies that $\cos \theta=\dfrac{\sqrt{7}}4.$ Now also note that $\tan \theta \neq \frac 92$ as stated in your problem.
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