Answer
$\dfrac{1+\tan^{2}u}{1-\tan^{2}u}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
Work Step by Step
$\dfrac{1+\tan^{2}u}{1-\tan^{2}u}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
On the left side of the equation, replace $\tan^{2}u$ by $\dfrac{\sin^{2}u}{\cos^{2}u}$:
$\dfrac{1+\dfrac{\sin^{2}u}{\cos^{2}u}}{1-\dfrac{\sin^{2}u}{\cos^{2}u}}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
Evaluate $1+\dfrac{\sin^{2}u}{\cos^{2}u}$ and $1-\dfrac{\sin^{2}u}{\cos^{2}u}$:
$\dfrac{\dfrac{\cos^{2}u+\sin^{2}u}{\cos^{2}u}}{\dfrac{\cos^{2}u-\sin^{2}u}{\cos^{2}u}}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
Evaluate the division on the left side:
$\dfrac{\sin^{2}u+\cos^{2}u}{\cos^{2}u-\sin^{2}u}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
Since $\sin^{2}u+\cos^{2}u=1$, the identity is proved:
$\dfrac{1}{\cos^{2}u-\sin^{2}u}=\dfrac{1}{\cos^{2}u-\sin^{2}u}$
