find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$

Question

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 ?

Answer 1

Because you have two different problems. Obviously (from your calculations which are correct) it can't be $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ at the same time.

Answer 2

This is an impossible situation. If you use the identity $$\tan(\theta) = {\sin(\theta)\over\cos(\theta)},$$ you get $\cos(\theta) = 1/6$.

However, $\cos(\theta) = \sqrt{7}/4$ by the Pythagorean identity.

Answer 3

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.