# How to Simplify Sin/tan problem.

I am trying to simplify $\displaystyle\frac{\sin^2}{\tan^2}$ but I don't know how to go about it. Any help is appreciated.

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You’re aware that $\tan=\sin/\cos$? – Lubin Apr 18 '14 at 3:39
@Lubin yes I am aware. – user2593789 Apr 18 '14 at 3:39
Then it’s just a matter of doing that substitution and simplifying the complicated fraction. – Lubin Apr 18 '14 at 3:40

Do you know that $$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$ You can rewrite this in terms of $\sin^2\theta$. $$\sin^2\theta=\tan^2\theta\cdot\cos^2\theta$$ The fraction $\dfrac{\sin^2\theta}{\tan^2\theta}$ can be rewritten as: $$\frac{\color{red}{\tan^2\theta}\cdot\cos^2\theta}{\color{red}{\tan^2\theta}}$$ Cancelling the $\tan^2\theta$ in the numerator and the denominator gives us $\cos^2\theta$ $$\displaystyle \color{green}{\therefore \frac{\sin^2\theta}{\tan^2\theta}=\cos^2\theta}$$
Hint: $$\frac{\sin^2}{\tan^2}=\frac{\sin^2}{\frac{\sin^2}{\cos^2}}$$ Can you simplify this fraction?