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

2 Answers
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!

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