Sin 2- -sin 2-cos 2- -cos 2- - - -

sin 2α #160; + sin 2βcos 2α cos 2β = cot (β α ) L.H.S= sin 2α + sin 2βcos2α #160; cos2β = 2sin ( 2α +2β2 )cos ( 2α 2β2 )2 sin ( 2α + ...

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Polar Representation of a Complex Number

sin 2α +sin 2...

Question

$\frac{sin 2 α + sin 2 β}{cos 2 α - cos 2 β} = cot (β - α)$

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Solution

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\frac{sin 2 α + sin 2 β}{cos 2 α - cos 2 β} = cot (β - α)

2 t a n^{2} α t a n^{2} β t a n^{2} γ + t a n^{2} α t a n^{2} β + t a n^{2} β t a n^{2} γ + t a n^{2} γ t a n^{2} α = 1

α, β, γ \neq (2 n + 1) \frac{π}{2}, \neq n π, n ϵ I

α, β

γ

cos 2 α + cos 2 β + cos 2 γ + {sin}^{2} α + {sin}^{2} β + {sin}^{2} γ =

c o s^{2} α - s i n^{2} α = t a n^{2} β .

t a n^{2} α = c o s^{2} β - s i n^{2} β .

cos α + cos β + cos α = 0 = sin α + sin β + sin α

{cos}^{2} α + {cos}^{2} β + {cos}^{2} α = \frac{3}{α} = {sin}^{2} α + {sin}^{2} β + {sin}^{2} α

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