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Question
If (cosec θ – sin θ) = a3 and (sec θ – cos θ) = b3, prove that a2b2(a2 + b2) = 1.
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Solution
We have `( cosec theta - sin theta ) = a^3`
= > ` a^3 = (1/ sin theta - sin theta)`
= > `a^3 = ((1- sin^2 theta))/sin theta = cos^2 theta / sin theta`
∴ `a=(cos^(2/3) theta)/(sin ^(1/3) theta)`
Again, `(sec theta - cos theta ) = b^3`
= >`b^3 = (1/cos theta - cos theta )`
=` ((1-cos^2 theta))/ cos theta`
=` (sin^2 theta)/cos theta`
∴ b =` (sin ^(2/3) theta)/(cos ^(1/3) theta)`
Now , LHS = `a^2 b^2 (a^2 + b^2 ) `
=` a^3 (ab^2) + ( a^2 b^2 ) b^3 `
=`a^3 ( ab^2 ) + ( a^2 b^2 ) b^3 `
=`(cos^2 theta)/(sin theta) xx [(cos ^(2/3) theta)/(sin^(1/3) theta) xx (sin ^(4/3)theta)/(cos ^(2/3) theta)] + [ ( cos ^(4/3) theta theta)/(sin ^(2/3) theta)xx(sin^(2/3)theta)/(cos ^(1/3)theta)] xx sin^2 theta/ cos theta`
=`cos^2 theta / sin theta xx sin theta + cos theta xx sin^2theta / costheta`
=`cos^2 theta + sin^2 theta = 1`
= RHS
Hence, proved
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