How do you show that (sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x?

Dec 18, 2016

The identity is false.

Explanation:

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

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

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

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

(sin^2x -cos^2x)/(cos^4xsin^4x) = ((sin^2x+ cos^2x)(sin^2x - cos^2x)/(sin^2xcos^2x)

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

As you can see, the two sides are unequal, so the identity is false.

Hopefully this helps!