Question

If tan A = √2 – 1, show that sin A cos A = root2/4

Answer 1

tanA=√2-1
or, tanA=(√2-1)(√2+1)/(√2+1)
or, tanA=(2-1)/(√2-1)
or, tanA=1/√2-1
∴, tanA+1/tanA=1/√2-1+√2-1
or, (tan²A+1)/tanA=[1+(√2-1)²]/(√2-1)
or, sec²A/tanA=(1+2-2√2+1)/(√2-1)
or, (1/cos²A)/(sinA/cosA)=(4-2√2)/(√2-1)
or, 1/sinAcosA=(4-2√2)(√2+1)/(√2-1)(√2+1)
or, 1/sinAcosA=(4√2-4+4-2√2)/(2-1)
or, 1/sinAcosA=2√2
or, sinAcosA=1/2√2
or, sinAcosA=√2/2√2.√2
or, sinAcosA=√2/4 (Proved)

Answer 2

 Given   Hence BC = √2 – 1, AB = 1 In right angled ΔABC, AC2 = AB2 + BC 2 [By Pythagoras theorem]⇒ AC2 = 12 + (√2 – 1)2 ⇒ AC2 = 1 + + 2 – 2√2 + 1 = 4– 2√2 

Hope this helps