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I need to proof this identity $$ 1 + \tan(\alpha)^2 = \frac {1}{cos(\alpha)^2} $$

I tried this: $$ \sin(\alpha)^2 + \cos(\alpha)^2 + \frac {\sin(\alpha)^2}{\cos(\alpha)^2} = \frac {1}{\cos(\alpha)^2} $$ However this leads me to nowhere. I also tried: $$ 1+\tan(\alpha)^2 = \frac {\sin(\alpha)^2 + \cos(\alpha)^2}{\cos(\alpha)^2} $$ This also lead me to nowhere. Any help is appreciated.

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  • $\begingroup$ The last line basically solves it. It didn't lead you nowhere. Just rewrite the last fraction as two fractions. Then what do you get? $\endgroup$ – Zelzy Mar 8 '16 at 17:25
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Hint:

$$\sin^2\alpha+\cos^2\alpha=1\stackrel{\cos\alpha\neq 0}\iff \frac{\sin^2\alpha}{\cos^2\alpha}+\frac{\cos^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha}$$

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    $\begingroup$ Got it thanks, I'll give you "best answer" when I can. $\endgroup$ – user218096 Feb 21 '15 at 7:57
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    $\begingroup$ This double implication is not true: the cases where $\cos(\alpha)=0$ must be excluded. Notice that here it was necessary to multiply, not to divide, by $\cos^2(\alpha)$. $\endgroup$ – MattAllegro Feb 22 '15 at 19:38
  • $\begingroup$ You should still make a note about the case of $\cos(\alpha)=0$ $\endgroup$ – graydad Mar 29 '15 at 18:25
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Also note that you can reach the result from where you started too: $$1+\frac {\sin^2 \alpha}{\cos^2 \alpha}=\frac{1}{\cos^2 \alpha} \iff 1=\frac{1-\sin^2 \alpha}{\cos^2 \alpha}=\frac{\cos^2\alpha}{\cos^2 \alpha}.$$

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  • $\begingroup$ I was unable to understand you, I got what you typed whoever $$ \frac {\cos(\alpha)^2} {\cos(\alpha)^2} = 1$$ right? i need $$\frac {1}{\cos(\alpha)^2}$$ $\endgroup$ – user218096 Feb 21 '15 at 8:08
  • $\begingroup$ We wanted to prove $$1+\frac{\sin^2 \alpha}{\cos^2 \alpha} =\frac{1}{\cos^2 \alpha}.$$ What I did is subtract from both sides $\frac{\sin^2 \alpha}{\cos^2 \alpha}$, which gave us the equivalent $$1=\frac{1-\sin^2\alpha}{\cos^2 \alpha}.$$ The last equality is true, because we know $1-\sin^2 \alpha = \cos^2 \alpha$, so the original equality is true as well. Have I made myself clear? $\endgroup$ – Vincenzo Oliva Feb 21 '15 at 8:15
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If you multiply $$1 + \tan^2(\alpha) = \frac {1}{\cos^2(\alpha)},$$ this is $$1 + \frac {\sin^2(\alpha)}{\cos^2(\alpha)} = \frac {1}{\cos^2(\alpha)},$$ by $\cos^2(\alpha)$, then you go straight to $$\cos^2(\alpha) + \sin^2(\alpha) = 1,$$ which is well known to be true for all $\alpha\in\mathbb{R}$.

(By the way, what's the most natural thing to do if you see $\cos^2(\alpha)$ is the only denominator?)

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So we have $$1 + \tan(\alpha)^2 = \frac {1}{cos(\alpha)^2}$$

Convert $\tan(\alpha)^2$ to $\frac {\sin(\alpha)^2}{\cos(\alpha)^2}$, so you have: $$1 + \frac {\sin(\alpha)^2}{\cos(\alpha)^2}$$ Next, you should make it one single fraction unit, so essentially you get the LCM.

Work that out, and you should get $$\frac {\cos(\alpha)^2 + \sin(\alpha)^2}{\cos(\alpha)^2 }$$

$\cos(\alpha)^2 + \sin(\alpha)^2 = 1$, so you get $$ \frac {1}{\cos(\alpha)^2 }$$

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