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Question
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
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Solution
Given that,
`cosec theta - sin theta = a^3` .....(1)
`sec theta - cos theta = b^3` ......(2)
We have to prove `a^2b^2(a^2 + b^2) = 1`
We know that `sin^2 theta + cos^2 theta = 1`
Now from the first equation, we have
`cosec theta - sin theta = a^3`
`=> 1/sin theta - sin theta = a^3`
`=> (1 - sin^2 theta)/sin theta = a^3`
`=> cos^2 theta/sin theta = a^3`
`=> a = (cos^(2/3) theta)/(sin^(1/3) theta)`
Again from the second equation, we have
`sec theta - cos theta =- b^3`
`=> 1/cos theta - cos theta = b^3`
`=> (1 - cos^2 theta)/cos theta = b^3`
`=> sin^2 theta/cos theta = b^3`
`=> b = (sin^(2/3) theta)/(cos^(1/3) theta)`
Therefore, we have
`a^2b^2 (a^2 + b^2) = (cos^(4/3) theta)/(sin^(2/3) theta cos^(2/3) theta) ((cos^(4/3) theta)/(sin^(2/3) theta) + (sin^(4/3) theta)/(cos^(2/3) theta))`
`= sin^(2/3) theta cos^(2/3) ((cos^(4/3) theta)/(sin^(2/3) theta) + (sin^(4/3) theta)/(cos^(2/3) theta))`
`= cos^(2/3) theta cos^(4/3) theta + sin^(2/3) theta sin^(4/3) theta`
`= cos^2 theta + sin^2 theta`
= 1
Hence proved.
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