Question

if tan A = √2-1 ,then prove that sinA cos A =1/2√2

Answer 1

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Answer 2

tan A = √2-1
(tan A = perpendicular/base)
so
perpendicular = √2-1
base = 1

according to Pythagoras thereom
in ∆ABC
AC² = AB²+BC²
AC² = (√2-1)²+(1)²
AC² = (√2)²+(1)²-2(√2)(1)+1
AC² = 2+1-2√2+1
AC² = 4-2√2
AC = √(4-2√2)

SO TANGENT = AC = √(4+2√2)

SIN A = PERPENDICULAR/TANGENT
SIN A = √2-1/√(4-2√2)


COS A = BASE/TANGENT
COS A = 1/√(4-2√2)

SO

4 = √16
2√2 = √(4×2) = √8

SIN A×COS A = 1/2√2

LHS = SIN A× COS A
LHS = √2-1/√(4-2√2) × 1/√(4-2√2)
LHS = √2-1/(√(4-2√2)².
LHS = √2-1/4-2√2
LHS = √2-1/√16+√8.
LHS = √2-1/√8(√2-1)

[CANCEL (√2-1)]

LHS = 1/√8
LHS = 1/√(4×2)
LHS = 1/2√2

LHS = RHS