To prove the identity [tex]\(\frac{1}{2}(\cot x + \tan x) = \csc 2x\)[/tex], let's work through it step-by-step:
### Step 1: Express the Left-Hand Side in terms of Sine and Cosine
The left-hand side of the identity is given as:
[tex]\[ \frac{1}{2}(\cot x + \tan x) \][/tex]
Using the trigonometric definitions:
[tex]\[ \cot x = \frac{\cos x}{\sin x} \][/tex]
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
Substitute these into the left-hand side:
[tex]\[ \frac{1}{2} \left( \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right) \][/tex]
### Step 2: Combine the Fractions
Find a common denominator to combine the fractions:
[tex]\[ \frac{1}{2} \left( \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right) \][/tex]
We know from the Pythagorean identity:
[tex]\[ \cos^2 x + \sin^2 x = 1 \][/tex]
Substitute this into the equation:
[tex]\[ \frac{1}{2} \left( \frac{1}{\sin x \cos x} \right) \][/tex]
### Step 3: Simplify the Fraction
[tex]\[ \frac{1}{2} \cdot \frac{1}{\sin x \cos x} = \frac{1}{2 \sin x \cos x} \][/tex]
### Step 4: Express in terms of [tex]\(\csc 2x\)[/tex]
We use the double-angle identity for sine:
[tex]\[ \sin 2x = 2 \sin x \cos x \][/tex]
Hence:
[tex]\[ \csc 2x = \frac{1}{\sin 2x} = \frac{1}{2 \sin x \cos x} \][/tex]
### Final Equality
So we have:
[tex]\[ \frac{1}{2}(\cot x + \tan x) = \frac{1}{2 \sin x \cos x} = \csc 2x \][/tex]
The identity we set out to prove:
[tex]\[ \frac{1}{2}(\cot x + \tan x) = \csc 2x \][/tex]
Looks like:
[tex]\[ \frac{1}{2}(\cot x+ \tan x) \neq \csc 2 x \][/tex]
Since the given identity does not hold under the rules we used and results in False.
