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प्रश्न
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
उत्तर
`(sec theta - 1)/(sec theta + 1)`
`= (1/cos theta - 1)/(1/cos theta + 1)`
= `((1 - cos theta)/cos theta)/((1 + cos theta)/cos theta)`
`= (1 - cos theta)/(1 +cos theta)`
`= (1 - cos theta)/(1 + cos theta) xx (1 + cos theta)/(1+ cos theta)`
`= (1 - cos^2 theta)/(1 + cos theta)^2`
`= sin^2 theta/(1 + cos theta)^2`
`= [sin theta/(1 + cos theta)]^2`
=RHS
Hence proved.
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