If secθ = 17/8, verify that

We have, secθ = (17k)/(8k) = AB/AC (For some value of k)

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

∴AB² = BC² + AC²

= (17k)² = BC² + (8k)²

= 289k² = BC² + 64k²

= BC² = 225k²

→ BC = 15k

∴ sinθ = BC/AB = (15k)/(17k) = 15/17

cosθ = AC/AB = (8k)/(17k) = 8/17

tanθ = BC/AC = sinθ /cosθ = 15/8

consider the LHS =

=

=

=

= ( - 33)/( - 611)

= 33/611

Now consider RHS =

=

=

= ( - 33)/( - 611)

= 33/611

∴ RHS = LHS

HENCE PROVED