# How do you verify (cotx/sec^2x) + (cotx/csc^2x)= cotx?

Nov 3, 2015

Verify trig expression.
$\left(\cot \frac{x}{\sec} ^ 2 x\right) + \left(\cot \frac{x}{\csc} ^ 2 x\right) = \cot x$

#### Explanation:

First term of left side:
$\left(\frac{\cos \frac{x}{\sin x}}{\frac{1}{{\cos}^{2} x}}\right) = \frac{{\cos}^{3} x}{\sin x}$ (1)

Second term of left side:
$\frac{\frac{\cos x}{\sin} x}{\frac{1}{\sin} ^ 2 x} = \frac{\cos x . {\sin}^{2} x}{\sin x}$ (2).

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

Nov 3, 2015

Determination is done by following method.

#### Explanation:

LHS= $\cot \frac{x}{\sec} ^ 2 x + \cot \frac{x}{\csc} ^ 2 x$

$\text{ } \textcolor{w h i t e}{a} = \cot x \left(\frac{1}{\sec} ^ 2 x + \frac{1}{\csc} ^ 2 x\right)$

$\text{ } \textcolor{w h i t e}{a} = \cot x \left({\cos}^{2} x + {\sin}^{2} x\right)$

$\text{ } \textcolor{w h i t e}{a} = \cot x$

$\text{ } \textcolor{w h i t e}{a} = R H S$