Prove the following identities.

(1 + cotθ – cosec θ)(1 + tanθ + secθ) = 2

Consider LHS,

LHS = (1 + cotθ – cosecθ) (1 + tanθ + secθ)

Expanding the above,

⇒ (1 + cotθ – cosecθ) (1 + tanθ + secθ)

= 1 + tanθ + secθ + cotθ + cotθtanθ + cotθsecθ – cosecθ –cosecθtanθ –cosecθsecθ

We know that = secθ, = cosecθ, = tanθ and = cotθ.

= 1 + tanθ + secθ + cotθ + 1 + cosecθ – cosecθ – secθ – cosecθsecθ

We know that tanθ + cotθ = cosecθsecθ.

= 1 + cosecθsecθ – cosecθsecθ + 1

∴ (1 + cotθ – cosecθ) (1 + tanθ + secθ) = 2 = RHS

Hence proved.