Question

if cosec theta - cot theta = root 2 cot theta, then prove that cosec theta + cot theta = root 2 cosec theta.

Answer 1

Answer:

We have,

$cosec \theta - cot \theta = \sqrt{2} cot \theta$

$\implies cosec \theta = \sqrt{2} cot \theta + cot \theta$

$\implies cosec \theta = cot \theta (\sqrt{2} + 1)$

$\implies \frac{cosec \theta }{\sqrt{2} + 1}=cot \theta$

$\implies \frac{cosec \theta \times (\sqrt{2} - 1)}{2 - 1}= cot \theta$

( By rationalizing )

$\implies \frac{ \sqrt{2} cosec \theta - cosec \theta}{1}=cot \theta$

$\implies \sqrt{2} cosec \theta - cosec \theta = cot\theta$

$\implies \sqrt{2} cosec \theta = cot\theta+cosec \theta$

Hence, proved.

Answer 2

Answer:

Here you go

Step-by-step explanation:

Given, cosec θ-cot θ √2 cot θ

Squaring both the sides,

cosec²θ + cot²θ - 2-cosec θ cot θ = 2cot²θ

or,

cosec²θ -cot²θ = 2cosec θ cot θ

[: a² - b² = (a + b)(a - b)

or, (cosec θ+ cot θ)(cosec θ-cot θ) = 2cosec θ cot θ

Given:

(cosec θ-cot θ = √2 cot θ)

or, cosec θ+ cotθ = 2cosecθcotθ/√2 cotθ

cosec θ+ cot θ = √2 cosec θ

Hence Proved.