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Trigonometry> If cot^2 theta = cot(theta-alpha) cot(the...

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If cot^2 theta = cot(theta-alpha) cot(theta-beta) show that cot 2theta = 1/2(cot alpha+ cot beta)

Harnoor , 6 Years ago

Grade 11

1 Answers

Bhagyashri verma

Last Activity: 6 Years ago

Hi,

Using the given equation, and using the formulas, we get,

$cot^2\Theta = cot(\Theta -\alpha ) cot(\Theta -\beta )$

$cot^2\Theta = (cot\Theta cot\alpha +1 / cot\Theta -cot\alpha )* ((cot\Theta cot\beta +1 )/ cot\Theta -cot\beta)$ $= (cot^2\Theta cot\alpha cot\beta+ cot\Theta cot\beta+ cot\Theta cot\alpha+1)/(cot\alpha cot\beta +cot^2\theta-cot\Theta cot\beta- cot\Theta cot\alpha)$

on rearranging,

$cot^4\theta -1 = cot\alpha cot\theta ( 1+ cot^2 \theta)+cot\theta cot\alpha (1+cot^2\theta)$

$cot^2\theta -1 = cot\theta (cot\alpha+ cot\beta)$

$(cot^2\theta -1)/cot\theta = (cot\alpha+ cot\beta)$

$cot2\theta = 1/2(cot\alpha+ cot\beta)$

In case to any difficulty, feel free to ask.

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