# Question #bca2e

Mar 28, 2017

It doesn't. A counterexample is $x = \frac{\pi}{6}$

#### Explanation:

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

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

Subtracting ${\cot}^{2} x$ from both sides,
${\sin}^{2} x = 1$

If, say, $x = \frac{\pi}{6}$ then ${\sin}^{2} x = \frac{1}{4} \ne 1$, and since we can provide a counterexample the identity is not true.

Plugging this counterexample into the original identity, we get ${\cot}^{2} \left(\frac{\pi}{6}\right) + {\sin}^{2} \left(\frac{\pi}{6}\right) = 3.25$ on the left side and ${\csc}^{2} \left(\frac{\pi}{6}\right) = 4$. Obviously $3.25 \ne 4$ so the identity cannot be true.

Mar 28, 2017

${\sin}^{2} x = 1$ in the end, hence both sides of the equation are accounted for.

#### Explanation:

${\cot}^{2} x + {\sin}^{2} x = {\csc}^{2} x$

Since ${\csc}^{2} x = 1 + {\cot}^{2} x$,
${\cot}^{2} x + {\sin}^{2} x = 1 + {\cot}^{2} x$
${\cot}^{2} x - {\cot}^{2} x + {\sin}^{2} x = 1$
${\sin}^{2} x = 1$

Therefore,
${\cot}^{2} x + {\sin}^{2} x = {\cot}^{2} x + 1 = {\csc}^{2} x$