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प्रश्न
`(cot ^theta)/((cosec theta+1)) + ((cosec theta + 1))/cot theta = 2 sec theta`
उत्तर
LHS = `cot theta/((cosec theta + 1)) + ((cosec theta + 1))/ cot theta`
=` (cot^2 theta + ( cosec theta +1)^2)/((cosec theta + 1) cot theta)`
=`(cot^2 theta + cosec ^2 theta + 2 cosec theta +1)/((cosec theta +1) cot theta)`
=` (cot^2 theta+ cosec ^2 theta + 2 cosec theta + cosec^2 theta - cot^2 theta)/((cosec theta +1) cot theta)`
=` (2 cosec ^2 theta + 2 cosec theta)/((cose theta +1) cot theta)`
=`(2 cosec theta ( cosec + 1))/((cosec theta + 1) cot theta)`
=`(2cosec theta)/(cot theta)`
=` 2 xx 1/sin thetaxx sin theta/ cos theta`
=`2 sec theta`
=RHS
Hence, LHS = RHS
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