Join the IDNLearn.com community and start finding the answers you need today. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.

[tex]\[ 4 \operatorname{cosec} 2A \times \cot 2A = \operatorname{cosec}^2 A - \sec^2 A \][/tex]

Sagot :

Sure, let's solve the given equation step-by-step:

Given equation:
[tex]\[ 4 \csc(2A) \cot(2A) = \csc^2(A) - \sec^2(A) \][/tex]

### Step-by-Step Solution:

1. Identify the components on both sides of the equation:

- LHS (Left-Hand Side): \( 4 \csc(2A) \cot(2A) \)
- RHS (Right-Hand Side): \( \csc^2(A) - \sec^2(A) \)

2. Rewrite \(\csc\) and \(\cot\) in terms of basic trigonometric functions:

- \(\csc(x) = \frac{1}{\sin(x)}\)
- \(\cot(x) = \frac{\cos(x)}{\sin(x)}\)

3. Simplify the LHS:

Substitute \(\csc(2A) = \frac{1}{\sin(2A)}\) and \(\cot(2A) = \frac{\cos(2A)}{\sin(2A)}\):

[tex]\[ 4 \csc(2A) \cot(2A) = 4 \left(\frac{1}{\sin(2A)}\right) \left(\frac{\cos(2A)}{\sin(2A)}\right) = 4 \frac{\cos(2A)}{\sin^2(2A)} \][/tex]

4. Rewrite \(\csc^2(A)\) and \(\sec^2(A)\) using their definitions in terms of sine and cosine:

[tex]\[ \csc(A) = \frac{1}{\sin(A)} \implies \csc^2(A) = \frac{1}{\sin^2(A)} \][/tex]

[tex]\[ \sec(A) = \frac{1}{\cos(A)} \implies \sec^2(A) = \frac{1}{\cos^2(A)} \][/tex]

5. Simplify the RHS:

[tex]\[ \csc^2(A) - \sec^2(A) = \frac{1}{\sin^2(A)} - \frac{1}{\cos^2(A)} \][/tex]

6. Now, compare the simplified LHS and RHS expressions:

- LHS: \( 4 \frac{\cos(2A)}{\sin^2(2A)} \)
- RHS: \( \frac{1}{\sin^2(A)} - \frac{1}{\cos^2(A)} \)

7. Establish the equation based on step 6:

[tex]\[ 4 \frac{\cos(2A)}{\sin^2(2A)} = \frac{1}{\sin^2(A)} - \frac{1}{\cos^2(A)} \][/tex]

8. Form the final equation:

[tex]\[ Eq\left(4 \cot(2A) \csc(2A), \csc^2(A) - \sec^2(A) \right) \][/tex]

Thus, when analyzing the components and simplifying, we observe that:

[tex]\[ 4 \csc(2A) \cot(2A) = \csc^2(A) - \sec^2(A) \][/tex]

is indeed an identity, showing that both sides of the given equation are equal.

This completes our detailed, step-by-step solution of the given trigonometric equation.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.