If sec+tan=m, then show that m2-1/m2+1=sin
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solution:
Given,
sec + tan= m
» 1/cos + Sin/cos =m
( taking LCM then cross multiplication)
1 + sin = m. cos
(1 + sin )^2 =m^2 . cos^2
(Squaring both sides)
We know that sin square theta + cos square theta equal to 1 that implies cos square theta equal to 1 minus sin square theta
=> (1 + sin ) ^2 = m ^2(1 - sin^2)
We know that a square minus b square equal to a + b into a minus b
=> cancelling (sin + 1) both sides
=> 1 + sin = m ^2 (1-sin)
» 1 + sin=m^2 - m^2 -m^2.sin
Rearranging terms ,
we get,
sin + m^2.sin= m ^2 - 1
sin(1+m^2) =m^2 -1
=> sin = m^2 - 1 /m^2+1
Thank you
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