If tanθ = 1/√7 , show that

We have, tanθ = k/(k√7) = BC/AC (For some value of k)

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

∴AB² = BC² + AC²

= AB² = (k)² + (k√7)²

= AB² = k² + 7k²

= 8k² = (2k√2)²

→ AB = 2k√2

∴ cosecθ = AB/BC = 2k√2

secθ = AB/AC = =

consider the LHS,

LHS =

=

=

= 48/64

= 3/4

= RHS

HENCE PROVED