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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.

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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)} $

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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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    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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    Thank Eric! It took me a while to see how to use that identity "*twice*" :)2012-05-07
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Hint :

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

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

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