How do you prove ${cot}^{2} (θ) + 5 = 6 cot (θ)$ $Cot^2 (theta) + 5 = 6 Cot (theta)$ ?

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How do you prove ${cot}^{2} (θ) + 5 = 6 cot (θ)$ $Cot^2 (theta) + 5 = 6 Cot (theta)$ ?

1 Answer

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Answer 1 · 2016-06-19T01:42:40

Impossible.

Explanation:

It is impossible. Take an example to prove it:
$t = \frac{π}{3}$ $t = pi/3$ --> $cot (\frac{π}{3}) = \frac{\sqrt{3}}{3}$ $cot (pi/3) = sqrt3/3$ --> ${cot}^{2} (\frac{π}{3}) = \frac{3}{9} = \frac{1}{3}$ $cot^2 (pi/3) = 3/9 = 1/3$
$\frac{1}{3} + 5$ $1/3 + 5$ different to $6 (\frac{\sqrt{3}}{3})$ $6(sqrt3/3)$
However, the equation is true when $t = \frac{π}{4} + 2 k π$ $t = pi/4 + 2kpi$
$t = \frac{π}{4}$ $t = pi/4$ --> cot t = 1 --> ${cot}^{2} t = 1$ $cot^2 t = 1$ -->
1 + 5 = 6(1). OK

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How do you prove ${cot}^{2} (θ) + 5 = 6 cot (θ)$ $Cot^2 (theta) + 5 = 6 Cot (theta)$ ?

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How do you prove Cot^2 (theta) + 5 = 6 Cot (theta)cot2(θ)+5=6cot(θ)Cot^2 (theta) + 5 = 6 Cot (theta)?

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How do you prove ${cot}^{2} (θ) + 5 = 6 cot (θ)$ $Cot^2 (theta) + 5 = 6 Cot (theta)$ ?