Question

prove this identity plzzzzzzz. full explanation

Answer 1

$\mathsf{Given :\;\dfrac{co{sec}\theta}{co{sec}\theta - 1} + \dfrac{co{sec}\theta}{co{sec}\theta + 1}}$

Taking LCM, We get :

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

★  We know that : (A + B)(A - B) = A² - B²

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

$\mathsf{\implies \dfrac{2co{sec}^2\theta}{co{sec}^2\theta - 1}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{co{sec}^2\theta - 1 = cot^2\theta}}}$

$\mathsf{\implies \dfrac{2co{sec}^2\theta}{cot^2\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

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

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

$\mathsf{\implies \dfrac{\dfrac{2}{sin^2\theta}}{\dfrac{cos^2\theta}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{2}{sin^2\theta} \times \dfrac{sin^2\theta}{cos^2\theta}}$

$\mathsf{\implies \dfrac{2}{cos^2\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\mathsf{\implies 2\;{sec}^2\theta}$