# Question 7dffd

Nov 2, 2014

Let us turn the left-hand side into the right-hand side of the equation.

${\tan}^{2} x - {\cot}^{2} x$

by $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$,

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

by finding the common denominator,

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

by factoring out the numerator,

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

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

={sin^2x-cos^2x}/{cos^2x sin^2x#

by splitting into two fractions,

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

by cancelling out common factors,

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

by $\sec x = \frac{1}{\cos} x$ and $\csc x = \frac{1}{\sin} x$,

$= {\sec}^{2} x - {\csc}^{2} x$

Hence,

${\tan}^{2} x - {\cot}^{2} x = {\sec}^{2} x - {\csc}^{2} x$.

I hope that this was helpful.