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If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$?

the answer is given as $-1$ in my book, but how do you arrive at this conclusion?

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    ....and the question could have been phrased as follows: Why is it that if $x + \frac 1 x = 1$ then $x^2 + \frac{1}{x^2} = -1$? No need to mention cotangents at that point in the problem.2012-08-28

4 Answers 4

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cota+ (1/cota)=1

Therefore, Squaring on both sides we get:

  cot^2a + (1/cot^2a)+ 2 = 1   Hence,    cot^2a + (1/cot^2a) = -1 
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Hint: $ x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2. $

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    Binomial formula: $(a+b)^2=a^2+2ab+b^2 \Rightarrow a^2+b^2=(a+b)^2-2ab.$2012-08-28
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Hint What do you get if you square $\cot(a)+\frac{1}{\cot(a)}$?

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Taking $x=\cot a$, $x+\frac{1}{x}=1\implies x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\frac{1}{x}=1-2=-1$

Alternatively, $x+\frac{1}{x}=1\implies x^2-x+1=0$

$x^2-x+1=0\implies x^3+1=0$

So,

$x^{3m}+(\frac{1}{x})^{3m}=(x^3)^m+\frac{1}{(x^3)^m}=2(-1)^m$

$x^{3m+1}+(\frac{1}{x})^{3m+1}=(x^3)^m\cdot x+\frac{1}{(x^3)^m\cdot x}=(-1)^m(x+\frac{1}{x})=(-1)^m$

$x^{3m+2}+(\frac{1}{x})^{3m+2}=(x^3)^m\cdot x^2+\frac{1}{(x^3)^m\cdot x^2}=(-1)^m(x^2+\frac{1}{x^2})=(-1)^m(-\frac{1}{x}-x)$ as $x^3=-1\implies x^2=-\frac{1}{x}$ and $\frac{1}{x^2}=x$

So,$x^{3m+2}+(\frac{1}{x})^{3m+2}=(-1)^{m+1}\cdot (x+\frac{1}{x})=(-1)^{m+1}$

If we put $m=0$, $x^2+\frac{1}{x^2}=(-1)^1=-1$

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    @aayush You don't even need to be aware of this formula. Just square $(x+\frac{1}{x})$.2012-08-28