In a right ΔABC, right - angled at B, it tanA = 1 then verify that 2sinA cosA = 1
Consider ΔABC to be a right - angled triangle.
tanA = 1 = BC/AB
→ BC = AB
Also, tanA = 1 = sinA/cosA
→ sinA = cosA
By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2
∴AC2 = BC2 + AB2
= AC2 = 2BC2
= (AC/BC)2 = 2
= AC/BC = √2
→ cosecA = √2
→ sinA = 1/√2 = cosA
2 sinA cosA = 2(1/√2)( 1/√2)
= 2(1/2)
= 1
∴ LHS = RHS
HENCE PROVED
33