# Solve $\frac{\tan^2x}{1+\tan^2x} = \sin^2x$

I'm having troubles with some problems here is an example : \begin{align} {\frac{\tan^2x}{1+\tan^2x}} & = \sin^2x \\ \end{align}

Can someone explain to me step by step on how you got the answer.

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Many of us can take it step by step. Can you show some effort? –  The Chaz 2.0 Nov 6 '12 at 17:55

All you need is:

• $\tan(x) = \dfrac{\sin(x)}{\cos(x)},$
• $\sin^2(x) + \cos^2(x) = 1$.

Try getting rid of the $\tan$'s first, then simplifying the fraction on the left a bit, and finally applying $\sin^2(x) + \cos^2(x) = 1$.

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The left hand side is $$\dfrac{\tan ^2 x}{\sec ^2 x} = \tan ^2 x \cdot \cos ^2 x$$

Can you take it from here?

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Joe, please do not edit my answer(s) to include reasoning that I omitted. If you would like to see an answer that is more thorough, or that spoon feeds the OP, feel free to write your own. I have rolled back to original. –  The Chaz 2.0 Nov 6 '12 at 18:51
The first time I ever saw the identity $$\tan(\alpha+\beta)=\text{a function of }\tan\alpha\text{ and }\tan\beta$$ it was proved by changing it to $$\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}$$ and then applying the previously demostrated identities involving those. The same thing works here: $$\frac{\tan^2 x}{1+\tan^2x} = \frac{\left(\frac{\sin^2 x}{\cos^2 x}\right)}{1+\frac{\sin^2 x}{\cos^2 x}}$$ Multiplying both the top and bottom by $\cos^2 x$ gets rid of the fractions-within-fractions and gives you $\cos^2 x+\sin^2 x$ in the denominator. You probably know that $\cos^2 x+\sin^2 x$ can be greatly simplified.