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I'm trying to prove the following identity:

$$\frac{\tan^2(x)+1}{\csc^2(x)} = \tan^2(x)$$

I tried this page but I couldn't make any sense out of their steps listed.

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  • $\begingroup$ There are quicker ways, but if you're stuck on something like this, just express everything in terms of $\sin$ and $\cos$ and try to simplify that. Then afterwards you can see what the trick was. $\endgroup$
    – dfan
    Feb 28, 2014 at 23:20
  • $\begingroup$ You can also multiply the numerator and denominator by $ \ \sin^2 x \ , $ (to get rid of the $ \ \csc^2 x \ $ "downstairs"), put the terms over a common denominator, factor the numerator, and apply the Pythagorean Identity. [Um, yeah, like TooTone did...] $\endgroup$ Feb 28, 2014 at 23:22
  • $\begingroup$ :) I wasn't reading the comments as I was writing my answer... I think you and dfan and I were all thinking on similar lines. What's remarkable is all the answers aren't exactly the same. $\endgroup$
    – TooTone
    Feb 28, 2014 at 23:30

6 Answers 6

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Recall that $\sec^2{x} = \tan^2{x} + 1$. So,

$$ \frac{\tan^2{x} + 1}{\csc^2{x}} = \frac{\sec^2{x}}{\csc^2{x}} = \frac{\sec^2{x}}{1} \cdot \frac{1}{\csc^2{x}} = \frac{1}{\cos^2{x}} \cdot \frac{\sin^2{x}}{1} = \frac{\sin^2{x}}{\cos^2{x}} = \tan^2{x}. $$

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  • $\begingroup$ Note that you could have gone directly to the third line instead of breaking apart $\tan^2 x + 1$ and then putting it back together. $\endgroup$
    – dfan
    Feb 28, 2014 at 23:26
  • $\begingroup$ @dfan Thanks for your suggestion. I streamlined it a bit. $\endgroup$ Feb 28, 2014 at 23:30
  • $\begingroup$ Thanks for breaking it down, your answer made the most sense to me. Separating the secant and cosecant like that is what I was missing. $\endgroup$
    – Tanner
    Feb 28, 2014 at 23:35
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If you start with the identity $\tan^2 x + 1 = \sec^2 x$, you get $\frac{\sec^2 x}{\csc^2 x}$. Now rewrite that in terms of $\sin x$ and $\cos x$ and you'll get your result.

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The way I'd choose is to simply convert everything into sines and cosines. So $\tan \rightarrow \sin/\cos$ and $\csc \rightarrow 1/\sin$:

$$\begin{align} (\tan^2x +1)\frac1{\csc^2x} &= \left( \frac{\sin^2x}{\cos^2x} + 1 \right)\sin^2 x \\ &= \left( \frac{\sin^2x + \cos^2x}{\cos^2x} \right)\sin^2 x \end{align}$$

hopefully you can see where to go from here?

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Using $\csc(x)=\frac1{\sin(x)}$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}$ you can write $$\frac{\tan^2(x)+1}{\csc^2(x)} = \sin^2(x)\cdot(\frac{\sin^2(x)}{cos^2(x)}+1) = \sin^2(x)\frac{\sin^2(x)+\cos^2(x)}{cos^2(x)}=\frac{\sin^2(x)\cdot1}{cos^2(x)}=\tan^2(x)$$

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$$\frac{\tan^2(x)+1}{\csc^2(x)} = \tan^2(x)\\ ({\tan^2(x)+1})\sin^2(x) = \tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}\\$$ Now, the RHS is $\frac{1}{\cos^2(x)}$, and the LHS is $$(\frac{\sin^2(x)}{\cos^2(x)}+1)=(\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)})=\frac{1}{\cos^2(x)}$$ Both are the same. Q.E.D.

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Another way , working the RHS $$ \frac{ \tan^2x + 1}{ \csc^2x} = \tan^2x $$ $$ \frac{ \tan^2x + 1}{ \csc^2x} = \tan^2x \frac{ \csc^2x}{ \csc^2x} $$ $$ \frac{ \tan^2x + 1}{ \csc^2x} = \frac{ \sec^2x}{ \csc^2x} $$

$$\frac{ \tan^2x + 1}{ \csc^2x} = \frac{ \tan^2x + 1}{ \csc^2x} $$

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