# How do you prove (1-sin^2theta)(1+cot^2theta)=cot^2theta?

Aug 4, 2018

#### Explanation:

We know that ,

$\left(1\right) {\cos}^{2} x + {\sin}^{2} x = 1$
$\left(2\right) {\csc}^{2} x - {\cot}^{2} x = 1$
$\left(3\right) \csc x = \frac{1}{\sin} x$
$\left(4\right) \cos \frac{x}{\sin} x = \cot x$

Using $\left(1\right) \mathmr{and} \left(2\right) :$

$L H S = \left(1 - {\sin}^{2} \theta\right) \left(1 + {\cot}^{2} \theta\right)$

$L H S = {\cos}^{2} \theta {\csc}^{2} \theta \to A p p l y \left(3\right)$

$L H S = {\cos}^{2} \theta \cdot \frac{1}{\sin} ^ 2 \theta$

$L H S = {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta \to A p p l y \left(4\right)$

$L H S = {\cot}^{2} \theta$

$L H S = R H S$

Aug 4, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

â€¢color(white)(x)cottheta=costheta/sintheta

$\text{consider the left side}$

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

$\text{expand the factors}$

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

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

$= 1 + {\cot}^{2} \theta - 1 \leftarrow {\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$= {\cot}^{2} \theta = \text{ right side "rArr" verified}$

Aug 5, 2018

As proved below.

#### Explanation:

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

$1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

$1 + {\cot}^{2} \theta = {\csc}^{2} \theta$

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

$\left(1 - {\sin}^{2} \theta\right) \left(1 + {\cot}^{2} \theta\right) = {\cos}^{2} \theta \cdot {\csc}^{2} \theta$

$\implies {\cos}^{2} \theta \cdot \frac{1}{\sin} ^ 2 \theta$

$\implies {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta = {\cot}^{2} \theta = R H S$