If a²sec²theta-b²tan²theta=c², then prove that sin²theta=c²-a/c²-b.
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a^2 sec^2 - b^2 tan^2 = c^2
a^2/cos^2 - b^2 sin^2/cos^2 = c^2
a^2 - b^2 sin^2 = c^2 cos^2
a^2 - b^2 sin^2 = C^2( 1 - sin^2)
a^2 - b^2 sin^2 = c^2 - c^2 sin^2
a^2 - c^2 =- c^2 sin^2 + b^2 sin^2
sin^2 = a^2 - c^2)/ b^2 - c^2
= c^2 - a^2)/ c^2 - b^2
a^2/cos^2 - b^2 sin^2/cos^2 = c^2
a^2 - b^2 sin^2 = c^2 cos^2
a^2 - b^2 sin^2 = C^2( 1 - sin^2)
a^2 - b^2 sin^2 = c^2 - c^2 sin^2
a^2 - c^2 =- c^2 sin^2 + b^2 sin^2
sin^2 = a^2 - c^2)/ b^2 - c^2
= c^2 - a^2)/ c^2 - b^2
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Hope this attachment will help you....
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but here is small error in given question
no...u r mistaken imao
anyways thanks for your help
try to understand it may be misprint in the book
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