Notes: Prove the trigonometric identity: (cot x)^2+1= (cosec x)^2.

Thursday, December 17, 2015

We'll manage the right side, knowing that the cosecant
function is given by the ratio:

$\displaystyle{\cos{{e}}}{c}{x}=\frac{{1}}{{{\sin{}}}}$
x)

We'll raise to square both
sides:

$\displaystyle{\cos{{e}}}{{c}}^{{2}}{x}=\frac{{1}}{{{{\sin}}^{{2}}}}$
x)

We'll re-write the
expression:

$\displaystyle{{\cot}}^{{2}}{x}+{1}=\frac{{1}}{{{{\sin}}^{{2}}}}$
x)

We'll multiply both sides by $\displaystyle{{\sin}}^{{2}}{x}$
:

$\displaystyle{{\sin}}^{{2}}{x}\cdot{{\cot}}^{{2}}{x}+{{\sin}}^{{2}}{x}=$
1

But $\displaystyle{{\cot}}^{{2}}{x}=\frac{{{{\cos}}^{{2}}{x}}}{{{{\sin}}^{{2}}}}$
x)

The expression will
become:

$\displaystyle\frac{{{{\sin}}^{{2}}{x}\cdot{{\cos}}^{{2}}{x}}}{{{{\sin}}^{{2}}{x}}}+{{\sin}}^{{2}}{x}=$
1

We'll simplify by $\displaystyle{{\sin}}^{{2}}{x}$ and we'll get the
Pythagorean identity:

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

Since the given expression led to the
Pythagorean identity, that means that the expression represents an identity,
too.

Notes