Question

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Answer 1

To prove:

$\longrightarrow \bf (cot \theta - cosec \theta)^2 = \dfrac {1 - cos \theta}{1 + cos \theta}$

Solution:

$\sf (cot \theta - cosec \theta)^2 = \dfrac {1 - cos \theta}{1 + cos \theta}$

$\bf LHS = (cot \theta - cosec \theta)^2$

$\longrightarrow \sf cot^2 \theta - 2 cot \theta \ cosec \theta + cosec^2 \theta$

$\longrightarrow \sf \dfrac {cos^2 \theta}{sin^2 \theta} - 2 \bigg( \dfrac{cos \theta}{sin \theta}\bigg) \bigg( \dfrac {1}{sin \theta} \bigg) + \dfrac {1}{sin^2 \theta}$

$\longrightarrow \sf \dfrac {cos^2 \theta}{sin^2 \theta} - 2 \dfrac {cos \theta}{sin^2 \theta} + \dfrac {1}{sin^2 \theta}$

$\longrightarrow \sf \dfrac {(cos^2 \theta - 2cos \theta + 1)}{sin^2 \theta}$

$\longrightarrow \sf \dfrac {(1 - cos \theta)^2}{1 - cos^2 \theta)}$

$\longrightarrow \sf \dfrac {(1 - cos \theta)^2}{(1 + cos \theta)(1 - cos \theta)}$

$\longrightarrow \bf \dfrac {1 - cos \theta}{1 + cos \theta}$

$\bf = RHS$

$\large \therefore \bf Hence \ Proved.$

Answer 2

Answer:

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