Question

Please anyone help me to do question no 20 in proper explanation

Answer 1

$\textsf{20(a).\;Given Equations are :}\\\\\\\sf{\bigstar\;\;\dfrac{x}{a}cos\theta + \dfrac{y}{b}sin\theta = 1\;--------\;(1)}\\\\\\\sf{\bigstar\;\;\dfrac{x}{a}sin\theta - \dfrac{y}{b}cos\theta = 1\;--------\;(2)}\\\\\\\textsf{Consider Equation (1) :}\\\\\\\sf{\implies \dfrac{x}{a}cos\theta + \dfrac{y}{b}sin\theta = 1}\\\\\\\textsf{Squaring on both sides, We get :}\\\\\\\sf{\implies \bigg(\dfrac{x}{a}cos\theta + \dfrac{y}{b}sin\theta\bigg)^2 = 1}$

$\sf{\implies \dfrac{x^2}{a^2}cos^2\theta + \dfrac{y^2}{b^2}sin^2\theta + 2\bigg(\dfrac{x}{a}cos\theta\bigg)\bigg(\dfrac{y}{b}sin\theta\bigg) = 1}\\\\\\\sf{\implies \dfrac{x^2}{a^2}cos^2\theta + \dfrac{y^2}{b^2}sin^2\theta + 2\bigg(\dfrac{xy}{ab}\bigg)(sin\theta cos\theta) = 1\;------\;(3)}$

$\textsf{Consider Equation (2) :}\\\\\\\sf{\implies \dfrac{x}{a}sin\theta - \dfrac{y}{b}cos\theta = 1}\\\\\\\textsf{Squaring on both sides, We get :}\\\\\\\sf{\implies \bigg(\dfrac{x}{a}sin\theta - \dfrac{y}{b}cos\theta\bigg)^2 = 1}\\\\\\\sf{\implies \dfrac{x^2}{a^2}sin^2\theta + \dfrac{y^2}{b^2}cos^2\theta - 2\bigg(\dfrac{x}{a}sin\theta\bigg)\bigg(\dfrac{y}{b}cos\theta\bigg) = 1}$

$\sf{\implies \dfrac{x^2}{a^2}sin^2\theta + \dfrac{y^2}{b^2}cos^2\theta - 2\bigg(\dfrac{xy}{ab}\bigg)(sin\theta cos\theta) = 1\;------\;(4)}$

$\textsf{Adding Equations (3) and (4), We get :}$

$\sf{\implies \dfrac{x^2}{a^2}cos^2\theta + \dfrac{y^2}{b^2}sin^2\theta + 2\bigg(\dfrac{xy}{ab}\bigg)(sin\theta cos\theta)+ \dfrac{x^2}{a^2}sin^2\theta + \dfrac{y^2}{b^2}cos^2\theta - 2\bigg(\dfrac{xy}{ab}\bigg)(sin\theta cos\theta) = 1 + 1}$

$\sf{\implies \dfrac{x^2}{a^2}cos^2\theta + \dfrac{y^2}{b^2}sin^2\theta + \dfrac{x^2}{a^2}sin^2\theta + \dfrac{y^2}{b^2}cos^2\theta = 2}\\\\\\\implies \dfrac{x^2}{a^2}(cos^2\theta + sin^2\theta) + \dfrac{y^2}{b^2}(cos^2\theta + sin^2\theta) = 2}\\\\\\We\;know\;that : \boxed{\bigstar\;\;\sf{sin^2\theta + cos^2\theta = 1}}\\\\\\\sf{\implies \dfrac{x^2}{a^2} + \dfrac{y^2}{b^} = 2}$

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$\sf{20(b).\;\;Given : \dfrac{1}{cosec\theta - cot\theta} - \dfrac{1}{sin\theta}}\\\\\\\sf{We\;know\;that : \boxed{\bigstar\;\;\sf{cosec^2\theta - cot^2\theta = 1}}$

$\sf{We\;know\;that : \boxed{\bigstar\;\;\sf{(a^2 - b^2) = (a + b)(a - b)}}$

$\sf{\implies cosec^2\theta - cot^2\theta = (cosec\theta + cot\theta)(cosec\theta - cot\theta) = 1}\\\\\\\implies \dfrac{1}{cosec\theta - cot\theta} = cosec\theta + cot\theta\\\\\\\textsf{Substituting the above value in the given question, We get :}\\\\\\\implies cosec\theta + cot\theta - \dfrac{1}{sin\theta}\\\\\\\sf{We\;know\;that : \boxed{\bigstar\;\;\sf{\dfrac{1}{sin\theta} = cosec\theta}}$

$\sf{\implies cosec\theta + cot\theta - cosec\theta}\\\\\\\implies cosec\theta - (cosec\theta - cot\theta)\\\\\\\ As:\;(cosec\theta + cot\theta)(cosec\theta - cot\theta) = 1\\\\\\ \implies (cosec\theta - cot\theta) = \dfrac{1}{cosec\theta + cot\theta}\\\\\\ \implies \dfrac{1}{sin\theta} - \dfrac{1}{cosec\theta + cot\theta}$