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प्रश्न
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
उत्तर
LHS = `(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+costheta)`
=`((1-cos theta)(1+ cos theta))/((1+ cos theta)(1+ cos theta)) {"Dividing the numerator and
denominator by "(1+ cos theta)}`
=`(1- cos^2 theta)/((1+ cos theta )^2)`
=`(sin^2 theta)/((1+ cos theta) ^2)`
= RHS
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