\(\rm \sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} + \sqrt {\frac{{1 - \cos \theta }}{{1 + \cos \theta }}} \) = .

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 2)

Answer 1

Given:

\(\rm \sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} + \sqrt {\frac{{1 - \cos \theta }}{{1 + \cos \theta }}} \)

Concept used:

1. sin θ = 1/cosec θ

2. sin² θ + cos² θ = 1

3. (a + b)(a - b) = a² - b²

Calculation:

\(\rm \sqrt {\frac{{1 + \cos θ }}{{1 - \cos θ }}} + \sqrt {\frac{{1 - \cos θ }}{{1 + \cos θ }}} \)

⇒ \(\sqrt {\frac{(1 + \cos θ)^2}{(1 - \cos θ)(1 + \cos θ)}} + \sqrt {\frac{(1 - \cos θ)^2}{(1 - \cos θ)(1 + \cos θ)}} \)

⇒ \(\sqrt { \frac{(1 + \cos θ)^2}{1 - \cos^2 θ} } + \sqrt {\frac{(1 - \cos θ)^2}{1 - \cos^2 θ}} \)

⇒ \(\sqrt { \frac{(1 + \cos θ)^2}{sin^2 θ} } + \sqrt {\frac{(1 - \cos θ)^2}{sin^2 θ}} \)

⇒ \({ \frac{(1 + \cos θ)}{sin θ} } + {\frac{(1 - \cos θ)}{sinθ}} \)

⇒ \({ \frac{(1 + \cos θ + 1 - \cos θ)}{sin θ} }\)

⇒ \({ \frac{2}{sin θ} }\)

⇒ 2 cosec θ

∴ The simplified answer is 2 cosec θ.

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