If cosθ = 3/5, show that
We have, cosθ = (3k)/(5k) = AC/AB (For some value of k)
By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2
∴AB2 = BC2 + AC2
= (5k)2 = BC2 + (3k)2
= 25k2 = BC2 + 9k2
= BC2 = 16k2
= (4k)2
→ BC = 4k
∴ sinθ = BC/AB = (4k)/(5k) = 4/5
tanθ = BC/AC = sinθ /cosθ = (4k)/(3k) = 4/3
cotθ = 1/tanθ = 3/4
consider the LHS
LHS = =
=
= 3/160
= RHS
HENCE PROVED