Question

sec - tane
1-2 sec 0 tan 0 + 2 tane
sec 0 + tan​

Answer 1

To prove:

$\rm \Longrightarrow \dfrac{sec \theta - tan \theta}{sec \theta + tan \theta} = 1 - 2 sec\theta \ tan\theta + 2 tan^2 \theta$

On solving the LHS we get:

$\rm \Longrightarrow \dfrac{sec \theta - tan \theta}{sec \theta + tan \theta}$

Multiply both the numerator and denominator by secθ - tanθ.

$\rm \Longrightarrow \dfrac{sec \theta - tan \theta}{sec \theta + tan \theta} \ \times \dfrac{sec \theta - tan \theta}{sec \theta - tan \theta}$

Identities used:

⇒ (a - b)(a - b) = (a - b)²

⇒ (a + b)(a - b) = a² - b²

$\rm \Longrightarrow \dfrac{\big(sec \theta - tan \theta\big)^2}{sec^2\theta - tan^2\theta}$

Identities used:

⇒ (a - b)² = a² + b² - 2ab

⇒ sec²θ - tan²θ = 1

$\rm \Longrightarrow \dfrac{sec^2\theta + tan^2\theta - 2sec\theta tan\theta}{1}$

Identities used:

⇒ sec²θ = 1 + tan²θ

$\rm \Longrightarrow \dfrac{1 + tan^2 \theta + tan^2\theta - 2sec\theta tan\theta}{1}$

$\rm \Longrightarrow \dfrac{1 + 2tan^2 \theta - 2sec\theta tan\theta}{1}$

$\rm \Longrightarrow 1 + 2tan^2 \theta - 2sec\theta tan\theta$

$\rm \Longrightarrow 1 - 2sec\theta tan\theta + 2tan^2 \theta$

LHS = RHS

Hence proved.