Given,

…….. (i)

∵ cos (90 – θ) = sin θ cos (90 – θ) = cos θ

sec (90 – θ) = cosec θ cot (90 – θ) = tan θ

cosec (90 – θ) = sec θ tan (90 – θ) = cot θ

Putting these values in (i),

We get,

∵ cosec θ = 1/sin θ and

sec θ = 1/cos θ

⇒

⇒ 1 + 1 = 2