# How does one verify \frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x ? \frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x I know that there are

How does one verify $\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{tan}}^{2}x}={\mathrm{cos}}^{2}x$ ?
$\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{tan}}^{2}x}={\mathrm{cos}}^{2}x$
I know that there are a few properties in here, but I don't know how the signs would effect them.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

sonorous9n
$\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{tan}}^{2}x}$
$=\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}$
$=\frac{\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)}{\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}$
$={\mathrm{cos}}^{2}x$ Proved.

Durst37
The given equation is,
$\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{tan}}^{2}x}={\mathrm{cos}}^{2}x$
LHS.
$⇒\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{1-\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}$
$⇒\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}$
$⇒\frac{\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right){\mathrm{cos}}^{2}x}{\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)}$
$⇒{\mathrm{cos}}^{2}x$
Hence Proved!