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