Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Many of us can take it step by step. Can you show some effort? –  The Chaz 2.0 Nov 6 '12 at 17:55

3 Answers 3

up vote 6 down vote accepted

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 $\sin^2(x) + \cos^2(x) = 1$.

share|improve this answer

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.

share|improve this answer
    
I saved this answer about 45 minutes ago, and it says here I answer two minutes ago. –  Michael Hardy Nov 6 '12 at 20:02
    
....and 30 seconds after I posted the comment above, it says I posted it "7 mins ago". –  Michael Hardy Nov 6 '12 at 20:03
    
....and then, a minute after that, it says my first comment above was "1 min ago". –  Michael Hardy Nov 6 '12 at 20:05
    
I never edited my answer here after posting it. I don't know what only the bottom half would appear. –  Michael Hardy Nov 7 '12 at 14:40

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?

share|improve this answer
4  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.