# Prove? cscx(1+cosx)(cscx-cotx)=1

Dec 30, 2016

See explanation

#### Explanation:

We will use the following:

• $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
• $\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$
• $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$
• $1 - {\cos}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$

With those,

$\csc \left(x\right) \left(1 + \cos \left(x\right)\right) \left(\csc \left(x\right) - \cot \left(x\right)\right)$

$= \left(\csc \left(x\right) + \cot \left(x\right)\right) \left(\csc \left(x\right) - \cot \left(x\right)\right)$

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

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

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

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

$= 1$

Dec 30, 2016

$L H S = \cos e c \theta \left(1 + \cos \theta\right) \left(\cos e c \theta - \cot \theta\right)$

$= \left(\cos e c \theta + \cos e c \theta \cos \theta\right) \left(\cos e c \theta - \cot \theta\right)$

$= \left(\cos e c \theta + \frac{1}{\sin} \theta \times \cos \theta\right) \left(\cos e c \theta - \cot \theta\right)$

$= \left(\cos e c \theta + \cot \theta\right) \left(\cos e c \theta - \cot \theta\right)$

$= \left(\cos e {c}^{2} \theta - {\cot}^{2} \theta\right) = 1 = R H S$

proved

See below:

#### Explanation:

We have:

$\csc x \left(1 + \cos x\right) \left(\csc x - \cot x\right) = 1$

distributing out:

$\left(\csc x + \csc x \cos x\right) \left(\csc x - \cot x\right) = 1$

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

and now to reorder and simplify:

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

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

Now we'll use the identity of ${\sin}^{2} x + {\cos}^{2} x = 1$ and so ${\sin}^{2} x = 1 - {\cos}^{2} x$

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

$1 = 1$