# Question de16d

Aug 6, 2016

We have: $\frac{{\cos}^{2} x - {\sin}^{2} x}{{\cot}^{2} x - {\tan}^{2} x}$

Start by writing tan and cot in terms of sin and cos

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

$= \frac{{\cos}^{2} x - {\sin}^{2} x}{\frac{{\cos}^{4} x - {\sin}^{4} x}{{\sin}^{2} x {\cos}^{2} x}} \cdot \frac{{\sin}^{2} x {\cos}^{2} x}{{\sin}^{2} x {\cos}^{2} x}$

$= \frac{{\cos}^{4} x {\sin}^{2} x - {\sin}^{4} x {\cos}^{2} x}{{\cos}^{4} x - {\sin}^{4} x}$

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

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

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

Aug 6, 2016

We have:
LHS=(cos^2x-sin^2x)/(cot^2x-tan^2#
$= \frac{{\sin}^{2} x {\cos}^{2} x \left({\cos}^{2} \frac{x}{{\sin}^{2} x {\cos}^{2} x} - {\sin}^{2} \frac{x}{{\sin}^{2} x {\cos}^{2} x}\right)}{{\csc}^{2} x - 1 - {\sec}^{2} x + 1}$

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

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