We have, cosecθ = 2 = 1/sinθ

→ sinθ = k/(2k) = BC/AB (For some value of k)

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

∴AB² = BC² + AC²

= (2k)² = (k)² + AC²

= 4k² = k² + AC²

= AC² = 3k²

→ AC = k√3

∴ cosθ = AC/AB = (k√3)/(2k) = √3/2

cotθ = cosθ /sinθ = AC/BC = √3

consider the LHS,

LHS = cotθ + = √3 +

= √ 3 +

=

=

=

= 2

= RHS

HENCE PROVED