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Class 10 > If tan|+sin|=m and tan|- sin|=n, show that (m2-n2)=4?mn
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If tan|+sin|=m and tan|- sin|=n, show that (m2-n2)=4?mn

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m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/
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m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX -tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now mn =(tanX + sinX).(tanX - sinX) mn =tan2X - sin2X = (sin2X/cos2X) - sin2X =sin2X((1/cos2X) -1) =sin2X((1-cos2X)/cos2X) =sin2X. (sin2X/ read less
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m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now, mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/cos2X)...
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m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2= tan2X + sin2X + 2tanX.sinX -tan2X - sin2X +2tanX.sinX m2-n2= 4tanX.sinX now, mn =(tanX + sinX).(tanX - sinX) mn =tan2X - sin2X = (sin2X/cos2X) - sin2X =sin2X((1/cos2X) -1) =sin2X((1-cos2X)/cos2X) =sin2X. (sin2X/cos2X) =sin2X.tan2X (mn)1/2 = sinX.tanX We know that, m2-n2= 4tanX.sinX now replacing "tanX.sinX", m2-n2= 4(mn)1/2 read less
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