Question
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
Solution
`[x/a cos theta + y/b sin theta]^2 + [x/a sin theta - y/b cos theta] = (1)^2 + (1)^2`
`x^2/a^2 cos^2 theta + y^2/b^2 sin^2 theta (2xy)/(ab) cos theta sin theta = x^2/a^2 sin^2 theta + y^2/b^2 cos^2 theta - (2xy)/(ab) sin theta cos theta = 1 + 1`
`x^2/a^2 cos^2 theta + y^2/b^2 cos^2 theta + y^2/b^2 sin^2 theta = 2`
`cos^ theta [x^2/a^2 + y^2/b^2] + sin^2 theta(x^2/a^2 + y^2/a^2) = 2`
`x^2/a^2 + y^2/b^2` = (∴ `cos^2 theta + sin^2 theta = 1`)
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Solution If X/A Cos Theta + Y/B Sin Theta = 1 and X/A Sin Theta - Y/B Cos Theta = 1 Prove that `X^2/A^2 + Y^2/B^2 = 2 Concept: Trigonometric Identities.
