Question

if tanA=√2-1 than proove sinAcosA=1/2√2

Answer 1

here, given tanA=√2-1
= 1/(√2 + 1) ................( by rationalising the denominator)
hence, we get 1/tanA = √2 + 1
so, tanA + 1/(tanA) = 2√2

therefore, now
LHS = (tan²A + 1)/tanA
= sec²A/tanA
= 1/sinAcosA
Hence, sinAcosA = 1/(2√2) ( proved)...

Answer 2

tanA = √2 -1
tanA = (√2-1).(√2+1)/(√2 +1)= 1/(√2 +1)
sinA = 1/√(4 + 2√2)
cosA = (√2 +1)/√(4+2√2)

so,

sinA .cosA = (√2+1)/(4+2√2)
= 1/2√2