Tabansi

2021-09-03

Proved the trigonometry identity
$\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}+{\mathrm{sec}}^{2}x+1}={\mathrm{cos}}^{2}x$

hajavaF

Given:
$\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}+{\mathrm{sec}}^{2}x+1}={\mathrm{cos}}^{2}x$
To prove:
The given identity.
Consider, left hand side of the given,
$\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}+{\mathrm{sec}}^{2}x+1}$
Using the trigonometric identity: ${\mathrm{cot}}^{2}x+1={\mathrm{csc}}^{2}x$
$⇒\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}x+{\mathrm{sec}}^{2}x+1}=\frac{{\mathrm{csc}}^{2}x}{{\mathrm{csc}}^{2}x+{\mathrm{sec}}^{2}x}$
Using:
${\mathrm{csc}}^{2}x=\frac{1}{{\mathrm{sin}}^{2}x}$
${\mathrm{sec}}^{2}x=\frac{1}{{\mathrm{cos}}^{2}x}$
$⇒\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}x+{\mathrm{sec}}^{2}x+1}=\frac{\left(\frac{1}{{\mathrm{sin}}^{2}x}\right)}{\left(\frac{1}{{\mathrm{sin}}^{2}x}+\frac{1}{{\mathrm{cos}}^{2}x}\right)}$
$=\frac{\left(\frac{1}{{\mathrm{sin}}^{2}x}\right)}{\left(\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}\right)}$
$=\frac{1}{{\mathrm{sin}}^{2}x}\cdot \frac{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}$
$=\frac{{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}$
Using the identity: ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$
$⇒\frac{{\mathrm{csc}}^{2}x}{{\mathrm{cot}}^{2}x+{\mathrm{sec}}^{2}x+1}=\frac{{\mathrm{cos}}^{2}x}{1}=1$
Hence the proof.

Jeffrey Jordon