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Question

If cosec x-sin x=a3, sec x-cos x=b3, then prove that a2 b2 a2+b2=1.

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Solution

cosec x-sin x=a31sin x-sin =a31-sin2xsin x=a3cos2xsin x=a3 a=cos2xsin x13 ....(i)Also, sec x-cos x=b31cos x-cos =b31-cos2xcos x=b3sin2xcos x=b3 b= sin2xcos x13 .....(ii)Now, LHS=a2b2a2+b2 =ab2a2+b2=cos2xsin x13sin2xcos x132 cos2xsin x132+sin2xcos x132=sin x cos x23 cos2x23sin x23+sin2x23cos x23=sin x cos x23cos3x23+sin3x23sin x23cos x23=sin x cos x23cos2x+sin2xsin x cos x23=1=RHS

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