# Prove (sin x - csc x)^2 = sin^2x + cot^2x - 1. Can anyone help me on this?

Apr 26, 2018

Show ${\left(\sin x - \csc x\right)}^{2}$$= {\sin}^{2} x + {\cot}^{2} x - 1$

#### Explanation:

${\left(\sin x - \csc x\right)}^{2}$

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

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

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

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

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

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

 = sin^2 x + cot^2 x - 1 quad sqrt

Apr 26, 2018

#### Explanation:

We need

$\csc x = \frac{1}{\sin} x$

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

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

Therefore,

$L H S = {\left(\sin x - \csc x\right)}^{2}$

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

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

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

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

$= R H S$

$Q E D$

Apr 26, 2018

Kindly find a Proof in the Explanation.

#### Explanation:

We will use the Identity : $\cos e {c}^{2} x = {\cot}^{2} x + 1$.

${\left(\sin x - \cos e c x\right)}^{2}$,

$= {\sin}^{2} x - 2 \sin x \cdot \cos e c x + \cos e {c}^{2} x$,

$= {\sin}^{2} x - 2 \sin x \cdot \frac{1}{\sin} x + {\cot}^{2} x + 1$,

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

$= {\sin}^{2} x + {\cot}^{2} x - 1$, as desired!