Question

please prove it please ​

Answer 1

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Answer 2

$\mathsf{Consider :\;\dfrac{1 - cos\theta}{1 + cos\theta}}$

Multiplying Numerator and Denominator with (1 - cosθ), We get :

$\mathsf{\implies \dfrac{(1 - cos\theta)(1 - cos\theta)}{(1 + cos\theta)(1 - cos\theta)}}$

★  We know that : (a + b)(a - b) = a² - b²

$\mathsf{\implies (1 - cos\theta)(1 + cos\theta) = 1 - cos^2\theta}$

$\mathsf{\implies \dfrac{(1 - cos\theta)^2}{1 - cos^2\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{1 - cos^2\theta = sin^2\theta}}}$

$\mathsf{\implies \dfrac{(1 - cos\theta)^2}{sin^2\theta}}$

$\mathsf{\implies \bigg[\dfrac{1 - cos\theta}{sin\theta}\bigg]^2}$

$\mathsf{\implies \bigg[\dfrac{1}{sin\theta} - \dfrac{cos\theta}{sin\theta}\bigg]^2}$

We know that :

$\bigstar\;\;\boxed{\mathsf{\dfrac{1}{sin\theta} = cosec\theta}}$

$\bigstar\;\;\boxed{\mathsf{\dfrac{cos\theta}{sin\theta} = cot\theta}}$

$\mathsf{\implies (cosec\theta - cot\theta)^2}$

$\mathsf{\implies [-(cot\theta - cosec\theta)]^2}$

$\mathsf{\implies [-1]^2(cot\theta - cosec\theta)^2}$

$\mathsf{\implies (cot\theta - cosec\theta)^2}$