Question

tanA=√2-1 prove sinAcosA=1/2√2

Answer 1

tanA=√2-1

LHS=sinA.cosA= 1/2(2sinA.cosA)

weknow ,
2sinA.cosA=sin2A

also , sin2A=2tanA/(1+tan^2A)
now,
= 1/2(2tanA/(1+tan^2A)

=tanA/(1+tan^2A)

=(√2-1)/{1+(√2-1)^2}

=(√2-1)/(1+3-2√2)

=(√2-1)/2√2(√2-1)

=1/2√2=RHS