InterviewSolution
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InterviewSolution
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If `cosec theta - sin theta = a^3 and sec theta - cos theta = b^3` then prove that `a^2b^2(a^2+b^2) = 1.` |
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Answer» `a^(3)=((1)/(sin theta)-sin theta)=((1-sin^(2)theta)/(sintheta))=(cos^(2)theta)/(sintheta) rArr a=(cos^(2//3)theta)/(sin^(1//3)theta).` `b^(3)=((1)/(cos theta)-cos theta)=((1-cos^(2)theta)/(costheta))=(sin^(2)theta)/(costheta) rArr b=(sin^(2//3)theta)/(cos^(1//3)theta).` `therefore a^(2)b^(2)(a^(2)+b^(2))=a^(4)b^(2)+a^(2)b^(4)=a^(3)(ab^(2))+(a^(2)b)b^(3)` `=(cos^(2)theta)/(sintheta)*[(cos^(2//3)theta)/(sin^(1//3)theta)*(sin^(4//3)theta)/(cos^(2//3)theta)]+[(cos^(4//3)theta)/(sin^(2//3)theta)*(sin^(2//3)theta)/(cos^(1//3)theta)]*(sin^(2)theta)/(costheta)` `=(cos^(2)theta)/(sintheta)*sin theta+cos theta*(sin^(2)theta)/(cos theta)=(cos^(2)theta+sin^(2)theta)=1.` |
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