# Prove that: 1+ sin 2x / sin^2 x = csc^2 x + 2cot x

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1 + sin2x / sin^2 x = csc^2 x + 2cot x

We will start from the right side.

We know that csc x = 1/sinx and cotx = cosx/sinx

==> csc^2 x + 2cotx = (1/sin^2 x) + 2 cosx/sinx

We will find the common denominator.

==> csc^2 x + 2cotx = (1+ 2sinxcosx)/sin^2 x

Now we know that 2sinx*cosx = sin2x

==> csc^2 x + 2cotx = (1+ sin2x)/sin^2 x ........q.e.d

**Then we proved that: csc^2 x + 2cotx = (1+ sin2x) /sin^2 x**

We have to prove that : (1+ sin 2x) / sin^2 x = csc^2 x + 2cot x

Start from the left hand side:

(1+ sin 2x) / sin^2 x

use the relation sin 2x = 2*sin x*cos x

=> (1 + 2*sin x*cos x) / (sin x)^2

=> 1/ (sin x)^2 + 2*sin x*cos x / (sin x)^2

we know 1/sin x = csc x and cos x / sin x = cot x

=> (csc x)^2 + 2*(cos x)/(sin x)

=> (csc x)^2 + 2*cot x

which is the right hand side

**This proves that (1+ sin 2x) / sin^2 x = csc^2 x + 2cot x**