If cotθ = 3/4, show that
We have, cotθ = (3k)/(4k) = AC/BC (For some value of k)
By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2
∴AB2 = BC2 + AC2
= AB2 = (4k)2 + (3k)2
= AB2 = 16k2 + 9k2
= AB2 = 25k2
= (5k)2
→ AB = 5k
∴ sinθ = BC/AB = (4k)/(5k) = 4/5
cosθ = AC/AB = (3k)/(5k) = 3/5
consider the LHS
LHS =
=
=
=
= 1/√7
= RHS
HENCE PROVED