Let's break down the given equation and simplify it using the provided trigonometric identities.

Given Equation:

$\frac{1}{2}(\cot x + \tan x) = \csc(2x)$

Step-by-Step Solution:

1. Express $\cot x$ and $\tan x$ in terms of $\sin x$ and $\cos x$:

Using the quotient identity:

• $\tan x = \frac{\sin x}{\cos x}$
• $\cot x = \frac{\cos x}{\sin x}$

Substitute these into the equation:

$\frac{1}{2} \left( \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right) = \csc(2x)$

2. Simplify the left-hand side (LHS):

The LHS becomes:

$\frac{1}{2} \left( \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right)$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, this simplifies to:

$\frac{1}{2} \left( \frac{1}{\sin x \cos x} \right)$

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

3. Apply the double-angle identity for sine:

$\sin(2x) = 2 \sin x \cos x$

Thus, the LHS becomes:

$\frac{1}{\sin(2x)} = \csc(2x)$

4. Conclusion:

The left-hand side equals the right-hand side of the equation:

$\frac{1}{\sin(2x)} = \csc(2x)$

This confirms that the equation is true.

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Summary of Steps:

1. Expressed $\cot x$ and $\tan x$ in terms of sine and cosine.
2. Simplified the expression using the Pythagorean identity.
3. Applied the double-angle identity for sine.
4. Confirmed that both sides of the equation are equal.

Would you like a more detailed explanation on any of these steps?

Related Questions:

1. How can you derive the Pythagorean identity from the unit circle?
2. How is the double-angle identity for sine related to the product of $\sin x$ and $\cos x$?
3. Can you solve for $x$ if given a specific value for $\csc(2x)$?
4. How would this equation change if $\cot x$ and $\tan x$ were expressed in terms of angles in different quadrants?
5. What is the graphical interpretation of the equation $\frac{1}{2}(\cot x + \tan x) = \csc(2x)$?

Tip:

When simplifying trigonometric equations, it's often helpful to convert everything into sine and cosine terms to apply identities easily.