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