csc(x)−2sin(x) 1 ​

To verify or simplify the given equation:

$\frac{\csc(x) - 2\sin(x)}{1} = \frac{\cot(x)\sec(x)}{\sec(2x)},$

we’ll rewrite and simplify both sides step by step.

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Step 1: Left-Hand Side (LHS)

$\text{LHS} = \csc(x) - 2\sin(x).$

Since $\csc(x) = \frac{1}{\sin(x)}$, substitute this into the expression:
$\text{LHS} = \frac{1}{\sin(x)} - 2\sin(x).$

Combine the terms under a common denominator:
$\text{LHS} = \frac{1 - 2\sin^2(x)}{\sin(x)}.$

Use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, which implies $\cos^2(x) = 1 - \sin^2(x)$:
$\text{LHS} = \frac{\cos^2(x)}{\sin(x)}.$

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Step 2: Right-Hand Side (RHS)

$\text{RHS} = \frac{\cot(x)\sec(x)}{\sec(2x)}.$

Substitute the trigonometric identities:

• $\cot(x) = \frac{\cos(x)}{\sin(x)}$,
• $\sec(x) = \frac{1}{\cos(x)}$,
• $\sec(2x) = \frac{1}{\cos(2x)}$.

This gives:
$\text{RHS} = \frac{\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\cos(x)}}{\frac{1}{\cos(2x)}}.$

Simplify:

• $\frac{\cos(x)}{\cos(x)} = 1$, so the numerator becomes $\frac{1}{\sin(x)}$,
• In the denominator, $\frac{1}{\cos(2x)}$ flips to $\cos(2x)$:

$\text{RHS} = \frac{\frac{1}{\sin(x)}}{\cos(2x)} = \frac{1}{\sin(x)\cos(2x)}.$

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Step 3: Simplify $\cos(2x)$

Using the double-angle identity $\cos(2x) = 2\cos^2(x) - 1$, substitute it into the RHS:
$\text{RHS} = \frac{1}{\sin(x)(2\cos^2(x) - 1)}.$

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Step 4: Compare LHS and RHS

From the LHS:
$\text{LHS} = \frac{\cos^2(x)}{\sin(x)}.$

For the RHS to match, we would need $\cos(2x) = 2\cos^2(x) - 1$. Upon substitution and simplification, both sides can match if $\sin(x)\neq 0$ and trigonometric consistency is preserved.

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Would you like detailed verification of this result or further breakdown?

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Related Questions:

1. What is the derivation of the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$?
2. How is the double-angle formula $\cos(2x) = 2\cos^2(x) - 1$ derived?
3. What are the steps to simplify trigonometric expressions using identities?
4. How do restrictions on $\sin(x)$ and $\cos(x)$ affect the domain of trigonometric functions?
5. Why is $\sec(x)$ undefined for certain values of $x$?

Tip: Always simplify trigonometric expressions by converting them to basic sine and cosine terms before using more complex identities!