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I am currently trying to solve the following equations : $$ (1-\cos^2 \theta)\cot^2 \theta = \cos^2 \theta$$ $\cos^2=\cos^2 $ I followed convential wisdom and started by only simplifiying one side of the eqaution (in this case the left) down. I ended up with this :

$$\cos^2 \theta \left(\frac{1-\cos^2 \theta} {\sin^2 \theta}\right)$$

(I got this through expanding and then factorising. I also tried various other methods.) However, this does not seem to work. I have solved these equations before, but this one is stumping me!

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    @DietrichBurde Possibly a typo of 'algebraic' – 2017-02-17
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    I see. Now it says "Trigonomtery". Last part "tery" reminds me on terry like Terry Tao. – 2017-02-17

3 Answers 3

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You seem like a beginner in trigonometry. First identity that you must know is \begin{align} \sin^2 x + \cos^2 x &= 1 \\\\ \end{align} Also you must know how to express tan, cot, cosec, sec in ratios of sin and cos. Here if you write cot as ratio of cos and sin and use the above identity for cos square term you will get your result (with some constraints on x)

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This is not an equation, but an identity (holding for $\theta$ not an integral multiple of $\pi$): $$ (1-\cos^2\theta)\cot^2\theta= \sin^2\theta\frac{\cos^2\theta}{\sin^2\theta}=\cos^2\theta $$

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Plenty of ways you can show that using simple trig identities. Like-

$ (1-\cos^2 \theta)\cot^2 \theta$

$ =((\sin^2\theta + \cos^2\theta) - \cos^2 \theta)\cot^2 \theta$

$ =(\sin^2\theta)(\cot^2 \theta)$

$ =(\sin^2\theta)\,\frac{\cos^2 \theta}{\sin^2 \theta}$

$= \cos^2 \theta$

But this is an identity, not an equation.

It works for all values of $\theta$ where $\sin\theta\ne 0$.

That is, when, $\theta\ne n\pi$ for any integer $n$.