Show that tanθ + cotθ = 2 Cosec2θ

Answer 1

We know that:

tanθ = sinθ/cosθ and cotθ = cosθ/sinθ

So,

tanθ + cotθ = sinθ/cosθ + cosθ/sinθ

Taking LCM,

(tanθ + cotθ) = (sin²θ + cos²θ)/(sinθcosθ)

(since, sin²θ + cos²θ = 1)

(tanθ + cotθ) = 1/(sinθcosθ)

Now, we know that:

cosecθ = 1/sinθ and secθ = 1/cosθ

So,

cosec²θ = (1/sinθ)² = 1/(sin²θ)

Also,

2sinθcosθ = sin2θ

(substitute sin2θ = 2sinθcosθ)

Therefore,

(tanθ + cotθ) = 1/(sinθcosθ) = 2/(2sinθcosθ)

(tanθ + cotθ) = 2/sin2θ = 2cosec²θ

Hence,

tanθ + cotθ = 2cosec²θ

Thus, we have proved that tanθ + cotθ = 2cosec²θ.

Answer 2

what a lot of work!
and you got it wrong, besides!

tanθ + cotθ
= sinθ/cosθ + cosθ/sinθ
= (sin^2 + cos^2)/(sinθ cosθ)
= 1/(sinθ cosθ)
= 2/(2sinθ cosθ)
= 2/sin2θ
= 2csc2θ

Answer 3

Thank you for pointing that out! I apologize for the error in my initial solution. You are correct that the final answer is 2csc2θ.