Question

Prove that: (cosec θ + cot θ)² = 1- cos θ÷1+cos θ​

Answer 1

Step-by-step explanation:

$\mathsf{Given :\;(cosec\theta + cot\theta)^2}$

$\textsf{We know that : \boxed{\mathsf{cosec\theta = \dfrac{1}{sin\theta}}}}$

$\textsf{We know that : \boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

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

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

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

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

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

$\textsf{We know that : \boxed{\mathsf{a^2 - b^2 = (a + b)(a - b)}}}$

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

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

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