Answer :
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.
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.
