We have, sinθ = (3k)/(4k) = BC/AB (For some value of k)

Consider the LHS

LHS = =

=

=

= cosθ /sinθ

= cotθ

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

∴AB² = BC² + AC²

= (4k)² = (3k)² + AC²

= 16k² = 9k² + AC²

= AC² = 7k²

→ AC = k√7

∴ cotθ = AC/BC = k/(k√7) = 1/√7

∴ LHS = RHS

HENCE PROVED