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$\begingroup$

I've been trying to prove it for a while, but can't seem to get anywhere.

$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$

Could someone please provide a valid proof?

I am not allowed to work on both sides of the equation.

Work so far:

RS:

$$ \begin{align} & \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt] & = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt] & = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)} \end{align} $$

I am completely lost after this.

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    Try to put everything on the same denominator.2012-05-07
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    @Raskolnikov I tried that but I end up getting an extremely messy and large equation. After that, I have absolutely no idea how to proceed.2012-05-07
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    What happens if you multiply the whole thing by $\sin^2\theta\cos^2\theta$?2012-05-07
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    Yes, but do it separately for each side of the equation, that's what I meant. (Or equivalently, what anon says.)2012-05-07
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    @Raskolnikov I am not allowed to work on both sides of the equation, because that apparently puts restrictions on the equation.2012-05-07
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    @anon My teacher doesn't allow us to work on both sides of the equation.2012-05-07
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    @AmitabhUdayiman I have tried expanding the right side of the equation and then turning everything into sines and cosines. I am lost after that.2012-05-07
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    Ok, then please post that mess here.The problem is not messy at all.That way you can get feedback on where you went wrong.2012-05-07
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    @AmitabhUdayiman Posted.2012-05-07
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    @ Farhad Yusufali Check the third and fourth terms in line 2 of what you have done.2012-05-07
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    the trigonometric identity values display or posted on internet are notclear solution to rearn on and are invaluable.im an engineering student from namibia and use to have ma self buzy with engineering mathematic books at ma place and enjoying trigonometric identity and equation,matric and relevantly matters2013-06-08

4 Answers 4

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Hint:

$$\tan(\theta) + \cot(\theta)= \frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}=\frac{\sin^2(\theta)}{\sin(\theta)\cos(\theta)}+\frac{\cos^2(\theta)}{\sin(\theta)\cos(\theta)} $$

12

Look at the largest triangle.

The

(There's a reason my avatar --the logo of my software company-- is a stylized version of this figure.)

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    Why have I never seen a diagram of a trig identity before?? (Besides the ones that should be memorized...) +12012-05-07
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    Very nice diagram, +1.2012-05-07
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    I like a good geometric proof without words ([for example](http://math.stackexchange.com/a/126075/)). (+1)2012-05-07
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Hint: $$(\tan\theta+\cot \theta)^2=\left(\frac{\sin\theta}{\cos \theta} +\frac{\cos \theta}{\sin \theta}\right)^2$$ $$= \left(\frac{\cos^2 \theta+\sin^2\theta}{\cos \theta \sin \theta}\right)^2.$$ Now try using the fact that $\cos^2\theta+\sin^2\theta=1$, twice.

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    $\frac{1}{(\cos^2\theta)(sin^2\theta)}$ $=\frac{1}{cos^2\theta} * \frac{1}{sin^2\theta}$ @EricNaslund Why I arrive at a wrong equation?2012-05-07
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    $$\dfrac{1}{\cos^2(\theta) \sin^2(\theta)} = \dfrac{\cos^2 (\theta) + \sin^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)} = \dfrac{\cos^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)} + \dfrac{\sin^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)}$$2012-05-07
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    @TheChaz Thank You! Just wondering but since I arrived at a multiplicative equation, does that mean $\frac{1}{\cos^2\theta} * \frac{1}{\sin^2\theta} = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta}$2012-05-07
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    You tell me! (If you add one more "$=$" to the end of my string of equalities, you'll have your result) :D2012-05-07
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    @TheChaz Thank You! :)2012-05-07
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    Thank Eric! It took me a while to see how to use that identity "*twice*" :)2012-05-07
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Hint :

$$\sin^2 \theta=\frac{\tan^2 \theta}{1+\tan^2 \theta}$$

$$\cos^2 \theta=\frac{1}{1+\tan^2 \theta}$$

$$\tan \theta = \frac{1}{\cot \theta}$$