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

Mar 13, 2018

See below.

#### Explanation:

I would rewrite in terms of sine and cosine.

${\cos}^{2} \frac{x}{\sin} ^ 2 x + \frac{1}{\sin} ^ 2 x = \frac{2}{\sin} ^ 2 x - 1$

$\frac{{\cos}^{2} x + 1}{\sin} ^ 2 x = \frac{2 - {\sin}^{2} x}{\sin} ^ 2 x$

Recall that ${\cos}^{2} x + {\sin}^{2} x = 1 \to {\sin}^{2} x = 1 - {\cos}^{2} x$.

$\frac{{\cos}^{2} x + 1}{\sin} ^ 2 x = \frac{2 - \left(1 - {\cos}^{2} x\right)}{\sin} ^ 2 x$

$\frac{{\cos}^{2} x + 1}{\sin} ^ 2 x = \frac{2 - 1 + {\cos}^{2} x}{\sin} ^ 2 x$

$\frac{{\cos}^{2} x + 1}{\sin} ^ 2 x = \frac{1 + {\cos}^{2} x}{\sin} ^ 2 x$

$L H S = R H S$

Proved!!

Hopefully this helps!

Mar 13, 2018

We know, ${\csc}^{2} x - {\cot}^{2} x = 1$

So,multiplying both side with $\left(- 1\right)$ we get,

${\cot}^{2} x - {\csc}^{2} x = - 1$

Now, adding $2 {\csc}^{2} x$ on both side of the equation we get,

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

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

Proved