How do you prove cot^2 x +csc^2 x = 2csc^2 x - 1?

2 Answers
Noah G
Mar 13, 2018

See below.

Explanation:

I would rewrite in terms of sine and cosine.

cos^2x/sin^2x+ 1/sin^2x= 2/sin^2x - 1

(cos^2x+ 1)/sin^2x = (2 - sin^2x)/sin^2x

Recall that cos^2x + sin^2x = 1 -> sin^2x =1 - cos^2x.

(cos^2x + 1)/sin^2x = (2 - (1 - cos^2x))/sin^2x

(cos^2x+ 1)/sin^2x = (2 - 1 + cos^2x)/sin^2x

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

LHS = RHS

Proved!!

Hopefully this helps!

Jane
Mar 13, 2018

We know, csc^2x -cot^2 x=1

So,multiplying both side with (-1) we get,

cot^2 x - csc^2x =-1

Now, adding 2 csc^2 x on both side of the equation we get,

cot^2 x - csc^2x + 2csc^2x=-1+2csc^2x

So,2 csc^2 x -1 = cot^2 x +csc^2x

Proved

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