Question

prove this please fast​

Answer 1

Answer:

refer to the above attachment for the solution

Answer 2

Proving $\frac{x^2}{a^2}+\frac{y^2}{b^2}\ =\ 2$

Step-by-step explanation:

$\ Given \ that:$

$\frac{x}{a}\ sin\theta -\frac{y}{b}\ cos\theta =1 \ \ \ and \ \ \frac{x}{a}\ cos\theta +\frac{y}{b}\ sin\theta =1$

$\ find: \\\\ \frac{x^2}{a^2}+ \frac{y^2}{b^2}\ = \ 2$

$\frac{x}{a}\ sin\theta -\frac{y}{b}\ cos\theta =1 \ ............(i) \\\\ \frac{x}{a}\ cos\theta +\frac{y}{b}\ sin\theta =1 \............(ii)$

$\ square \ the \ equation \ and \ then \ add$

equation ...........(i)

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

equation ...........(ii)

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

add both equation (i) + (ii)

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

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

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

proved..

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• Prove: https://brainly.in/question/8157059