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x/a cosA+ y/b sinA=1 x/a cosB + y/b sinB=1 cosAcosB/a^2 +sinAsinB/b^2=0 then show that tanAtanB=b^2(x^2-a^2)/a^2(y^2-b^2) and x^2+y^2=a^2-b^2

x/a cosA+ y/b sinA=1
x/a cosB + y/b sinB=1
cosAcosB/a^2 +sinAsinB/b^2=0 then show that tanAtanB=b^2(x^2-a^2)/a^2(y^2-b^2)  and x^2+y^2=a^2-b^2

Grade:11

1 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
2 years ago
Dear student

Please use brackets and repost so that question can be understood clearly

Regards

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