Question #bca2e

2 Answers
Mar 28, 2017

It doesn't. A counterexample is x=pi/6

Explanation:

One form of the Pythagorean Identity states that cot^2x+1=csc^2x

If cot^2x+sin^2x=csc^2x then by the transitive property, cot^2x+sin^2x=cot^2x+1

Subtracting cot^2x from both sides,
sin^2x=1

If, say, x=pi/6 then sin^2x=1/4!=1, and since we can provide a counterexample the identity is not true.

Plugging this counterexample into the original identity, we get cot^2(pi/6)+sin^2(pi/6)=3.25 on the left side and csc^2(pi/6)=4. Obviously 3.25!=4 so the identity cannot be true.

Mar 28, 2017

sin^2x=1 in the end, hence both sides of the equation are accounted for.
Read on for more info...

Explanation:

cot^2x+sin^2x=csc^2x

Since csc^2x=1+cot^2x,
cot^2x+sin^2x=1+cot^2x
cot^2x-cot^2x+sin^2x=1
sin^2x=1

Therefore,
cot^2x+sin^2x=cot^2x+1=csc^2x