Question

please solve this question​

Answer 1

Step-by-step explanation:

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Answer 2

Solution:

Given:

$\tt{\dfrac{x}{a}cos\theta + \dfrac{y}{b}sin\theta = 1}$                 ------[1]

And, $\tt{\dfrac{x}{a}sin\theta - \dfrac{y}{b}cos\theta = 1}$       ------[2]

To prove:

$\tt{\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 2}$

Proof:

On squaring the equation [1], we get

$\tt{( \dfrac{x}{a}cos\theta + \dfrac{y}{b}sin\theta)^{2} = 1^{2} }$

or $\tt{\dfrac{x^{2}}{a^{2}}cos^{2} \theta + \dfrac{y^{2}}{b^{2}}sin^{2} \theta + 2\dfrac{x}{a}\dfrac{y}{b} sin\theta cos\theta = 1}$    -------[3]

Also, on squaring the equation [2], we get

$\tt{( \dfrac{x}{a}sin\theta - \dfrac{y}{b} cos\theta)^{2} = 1^{2} }$

or $\tt{\dfrac{x^{2}}{a^{2}}sin^{2} \theta + \dfrac{y^{2}}{b^{2}}cos^{2} \theta - 2\dfrac{x}{a}\dfrac{y}{b} sin\theta cos\theta = 1}$    -------[4]

On adding equations [3] and [4], we get

$\tt{\dfrac{x^{2}}{a^{2}}cos^{2} \theta + \dfrac{y^{2}}{b^{2}}sin^{2} \theta + 2\dfrac{x}{a}\dfrac{y}{b} sin\theta cos\theta + \dfrac{x^{2}}{a^{2}}sin^{2} \theta + \dfrac{y^{2}}{b^{2}}cos^{2} \theta - 2\dfrac{x}{a}\dfrac{y}{b} sin\theta cos\theta = 1+1}$

or $\tt{\dfrac{x^{2}}{a^{2}}cos^{2} \theta + \dfrac{y^{2}}{b^{2}}sin^{2} \theta + \dfrac{x^{2}}{a^{2}}sin^{2} \theta + \dfrac{y^{2}}{b^{2}}cos^{2} \theta = 2}$

or $\tt{\dfrac{x^{2}}{a^{2}}(cos^{2} \theta +sin^{2} \theta) + \dfrac{y^{2}}{b^{2}}(sin^{2} \theta +cos^{2} \theta) = 2}$

or $\tt{\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 2}$

Hence Proved

Identities used:

(a+b)² = a² + b² + 2ab

(a-b)² = a² + b² - 2ab

sin²θ + cos²θ = 1