We have, secθ = 5/4 = 1/cosθ

→ cosθ = (4k)/(5k) = AC/AB (For some value of k)

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

∴AB² = BC² + AC²

= (5k)² = BC² + (4k)²

= 25k² = BC² + 16k²

= BC² = 9k²

→ BC = 3k

∴ sinθ = BC/AB = (3k)/(5k) = 3/5

consider the LHS

LHS = =

=

=

=

=

= 12/7

= RHS

HENCE PROVED