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