# How do you prove that 9sec^2u - 5tan^2u = 5 + 4sec^2u?

Feb 16, 2017

Convert to all sine and cosine using $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$.

$\frac{9}{\cos} ^ 2 u - \frac{5 {\sin}^{2} u}{\cos} ^ 2 u = 5 + \frac{4}{\cos} ^ 2 u$

$\frac{9 - 5 {\sin}^{2} u}{\cos} ^ 2 u = \frac{5 {\cos}^{2} u + 4}{\cos} ^ 2 u$

Now you should notice that the numerator on both sides have different trig functions. The one on left is in sine, while the one on right is in cosine. We can convert between the two using

$\textcolor{red}{{\sin}^{2} x + {\cos}^{2} x = 1}$

This means that ${\sin}^{2} x = 1 - {\cos}^{2} x$, so:

$\frac{9 - 5 \left(1 - {\cos}^{2} u\right)}{\cos} ^ 2 u = \frac{5 {\cos}^{2} u + 4}{\cos} ^ 2 u$

$\frac{9 - 5 + 5 {\cos}^{2} u}{\cos} ^ 2 u = \frac{5 {\cos}^{2} u + 4}{\cos} ^ 2 u$

$\frac{4 + 5 {\cos}^{2} u}{\cos} ^ 2 u = \frac{5 {\cos}^{2} u + 4}{\cos} ^ 2 u$

Since the $R H S = L H S$, this identity has been proved.

Hopefully this helps!