How do you verify the identity (cot^2x-1)/(1+cot^2x)=1-2sin^2x?

Sep 5, 2016

see explanation.

Explanation:

We attempt to express the left side in the same form as the right side.
Let's begin by rewriting ${\cot}^{2} x$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cot x = \frac{\cos x}{\sin x}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

rArr((cos^2x)/(sin^2x)-1)/(1+cos^2x/(sin^2x)

now multiply all terms on numerator and denominator by ${\sin}^{2} x$

$\Rightarrow \frac{{\cos}^{2} x - {\sin}^{2} x}{{\sin}^{2} x + {\cos}^{2} x} \ldots \ldots . \left(A\right)$

To simplify we require the $\textcolor{b l u e}{\text{trigonometric identities}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos 2 x = {\cos}^{2} x - {\sin}^{2} x = 1 - 2 {\sin}^{2} x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} x + {\cos}^{2} x = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Substituting these into (A) gives.

$\cos 2 x = 1 - 2 {\sin}^{2} x$

Thus left side = right side $\Rightarrow \text{ verified}$