Question

verify that the equation is an identity.
\frac{2}{1 + \cos x}-\tan ^{2} \frac{x}{2}=1
to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and transform the expression at each step.
\frac{2}{1 + \cos x}-\tan ^{2} \frac{x}{2}=\frac{2}{1 + \cos x}-\square
apply a pythagorean identity.
divide each term by tangent.
apply a reciprocal identity.
apply a half - angle identity.

Explanation:

Step1: Apply a Pythagorean identity

We know that \(\tan^{2}\frac{x}{2}=\frac{1 - \cos x}{1+\cos x}\) (derived from \(\tan^{2}\alpha=\frac{1-\cos2\alpha}{1 + \cos2\alpha}\) with \(\alpha=\frac{x}{2}\)).
Substitute \(\tan^{2}\frac{x}{2}\) into the left - hand side of the equation \(\frac{2}{1+\cos x}-\tan^{2}\frac{x}{2}\):

$$
LATEXBLOCK0
$$

Step2: Simplify the numerator

Simplify the numerator of \(\frac{2-(1 - \cos x)}{1+\cos x}\):

$$
LATEXBLOCK1
$$

So, \(\frac{2-(1 - \cos x)}{1+\cos x}=\frac{1+\cos x}{1+\cos x}\)

Step3: Simplify the fraction

Since \(\frac{1+\cos x}{1+\cos x}=1\) (for \(1+\cos x
eq0\), i.e., \(x
eq(2k + 1)\pi,k\in\mathbb{Z}\))

Answer:

The left - hand side \(\frac{2}{1+\cos x}-\tan^{2}\frac{x}{2}\) simplifies to \(1\), which is equal to the right - hand side. So, the equation \(\frac{2}{1+\cos x}-\tan^{2}\frac{x}{2}=1\) is an identity.