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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.

Gerry Myerson
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  • Try to put everything on the same denominator. – Raskolnikov May 07 '12 at 15:44
  • @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. –  May 07 '12 at 15:45
  • What happens if you multiply the whole thing by $\sin^2\theta\cos^2\theta$? – anon May 07 '12 at 15:47
  • Yes, but do it separately for each side of the equation, that's what I meant. (Or equivalently, what anon says.) – Raskolnikov May 07 '12 at 15:47
  • @Raskolnikov I am not allowed to work on both sides of the equation, because that apparently puts restrictions on the equation. –  May 07 '12 at 15:48
  • @anon My teacher doesn't allow us to work on both sides of the equation. –  May 07 '12 at 15:49
  • @AmitabhUdayiman I have tried expanding the right side of the equation and then turning everything into sines and cosines. I am lost after that. –  May 07 '12 at 15:49
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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. – Amitabh Udayiman May 07 '12 at 15:51
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    @AmitabhUdayiman Posted. –  May 07 '12 at 16:04
  • @ Farhad Yusufali Check the third and fourth terms in line 2 of what you have done. – Amitabh Udayiman May 07 '12 at 17:11
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4 Answers4

14

Look at the largest triangle.

The "Complete Triangle"

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

Blue
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7

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.

Eric Naslund
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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?

     –  May 07 '12 at 16:16
  • $$\dfrac{1}{\cos^2(\theta) \sin^2(\theta)} = \dfrac{\cos^2 (\theta) + \sin^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)} = \dfrac{\cos^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)} + \dfrac{\sin^2 (\theta)}{\cos^2(\theta) \sin^2(\theta)}$$ – The Chaz 2.0 May 07 '12 at 16:22
  • @TheChaz Thank You! Just wondering but since I arrived at a multiplicative equation, does that mean

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

     –  May 07 '12 at 16:25
  • You tell me! (If you add one more "$=$" to the end of my string of equalities, you'll have your result) :D – The Chaz 2.0 May 07 '12 at 16:31
  • @TheChaz Thank You! :) –  May 07 '12 at 16:37
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    Thank Eric! It took me a while to see how to use that identity "twice" :) – The Chaz 2.0 May 07 '12 at 16:37
1

Hint:

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

N. S.
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1

Hint :

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

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

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

Pedja
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