Consider LHS,

LHS = secθ + tanθ =

= (1)

We have, sinθ = (ak)/(bk) = BC/AB (For some value of k)

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

∴AB² = BC² + AC²

= (bk)² = (ak)² + AC²

= AC² = b²k² - a²k²

→ AC =

∴ cosθ = AC/AB = =

∴ from(1)

LHS = =

=

=

= RHS

HENCE PROVED