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If tanA+sinA=m and tanA-sinA=n, show that m2-n2 = 4√mn
Ans: TanA + SinA = m TanA - SinA = n.
m2-n2=4√mn.
m2-n2= (TanA + SinA)2-(TanA - SinA)2
= 4 TanA SinA
RHS 4√mn = 4
= 4 sin2A/cos2 A = 4Tan A sin A
m2-n2=4√mn
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