if x = a cos?, y = b sin?, then find b^2x^2 + a^2y^2 - a^2b^2

Asked by Renoy Gladson | 13th Jul, 2013, 03:35: PM

Expert Answer:

b^2x^2 + a^2y^2 - a^2b^2
b^2(acos?)^2 + a^2(bsin?)^2 - a^2b^2
= b^2 a^2 cos^2? + a^2b^2sin^2? - a^2b^2
= a^2b^2 ( cos^2? + sin^2?) - a^2b^2
Since,  cos^2? + sin^2? = 1
= a^2b^2 (1) - a^2b^2
= 0

Answered by  | 13th Jul, 2013, 07:12: PM

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