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Question

If xacos θ+ybsin θ=1and xasin θybcos θ=1,prove that x2a2+y2b2=2

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Solution

x over a cos theta plus y over b sin theta space equals 1 ---------- (i)
x over a sin theta minus y over b cos theta space equals 1 ---------- (ii)
Square both equations and add
equals x squared over a squared cos squared theta plus y squared over b squared sin squared theta space plus 2 space x over a cos theta space y over b sin theta plus space fraction numerator begin display style x squared end style over denominator begin display style a squared end style end fraction sin squared theta plus fraction numerator begin display style y squared end style over denominator begin display style b squared end style end fraction cos squared theta minus 2 x over a sin theta fraction numerator y over denominator b space end fraction cos theta equals x squared over a squared left parenthesis sin squared theta plus cos squared theta right parenthesis plus y squared over b squared left parenthesis sin squared theta plus cos squared theta right parenthesis space equals fraction numerator begin display style x squared end style over denominator begin display style a squared end style end fraction plus fraction numerator begin display style y squared end style over denominator begin display style b squared end style end fraction
LHS = RHS
Hence proved

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