m2+n2(m2−n2)
m2+n22mn
mn2(m2+n2)
m2n22(m2−n2)
Correct answer is B
Tan θ = m2−n22mn
OppAdj by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 = m2+n22mn
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