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Let's tackle the given equation step-by-step:
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = \tan^2 A + \cot 2A + 7 \][/tex]
### Step 1: Simplify the Left-Hand Side (LHS)
#### Part 1: Expand the terms
First, we need to expand each term on the LHS.
1. For [tex]\((\sin A + \csc A)^2\)[/tex]:
[tex]\[ (\sin A + \csc A)^2 = \sin^2 A + 2 \sin A \csc A + \csc^2 A \][/tex]
Since [tex]\(\csc A = \frac{1}{\sin A}\)[/tex]:
[tex]\[ = \sin^2 A + 2 \sin A \cdot \frac{1}{\sin A} + \left( \frac{1}{\sin A} \right)^2 = \sin^2 A + 2 + \csc^2 A = \sin^2 A + 2 + \frac{1}{\sin^2 A} = 1 - \cos^2 A + 2 + \frac{1}{\sin^2 A} \][/tex]
2. For [tex]\((\cos A + \sec A)^2\)[/tex]:
[tex]\[ (\cos A + \sec A)^2 = \cos^2 A + 2 \cos A \sec A + \sec^2 A \][/tex]
Since [tex]\(\sec A = \frac{1}{\cos A}\)[/tex]:
[tex]\[ = \cos^2 A + 2 \cos A \cdot \frac{1}{\cos A} + \left( \frac{1}{\cos A} \right)^2 = \cos^2 A + 2 + \sec^2 A = \cos^2 A + 2 + \frac{1}{\cos^2 A} = 1 - \sin^2 A + 2 + \frac{1}{\cos^2 A} \][/tex]
Now combine the results:
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = (1 - \cos^2 A + 2 + \frac{1}{\sin^2 A}) + (1 - \sin^2 A + 2 + \frac{1}{\cos^2 A}) = 4 + (1 - \cos^2 A + \frac{1}{\sin^2 A}) + (1 - \sin^2 A + \frac{1}{\cos^2 A}) \][/tex]
Further simplification shows:
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = \frac{10 \cos^2 A - 18}{\cos(4A) - 1} \][/tex]
### Step 2: Simplify the Right-Hand Side (RHS)
The RHS is:
[tex]\[ \tan^2 A + \cot 2A + 7 \][/tex]
### Step 3: Check if LHS = RHS
To verify the equality, we can check if the simplified forms for LHS and RHS are equivalent or not:
Given:
[tex]\[ \text{LHS} = \frac{10 \cos(2A)^2 - 18}{\cos(4A) - 1} \][/tex]
[tex]\[ \text{RHS} = \tan^2 A + \cot 2A + 7 \][/tex]
### Step 4: Solution
We compare the simplified forms:
[tex]\[ \frac{10 \cos(2A)^2 - 18}{\cos(4A) - 1} \neq \tan^2 A + \cot 2A + 7 \][/tex]
Thus, the given equation does not hold true for all values of [tex]\(A\)[/tex]. Therefore, the solution to the given equation is that the LHS does not equal the RHS under general assumptions.
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = \tan^2 A + \cot 2A + 7 \][/tex]
### Step 1: Simplify the Left-Hand Side (LHS)
#### Part 1: Expand the terms
First, we need to expand each term on the LHS.
1. For [tex]\((\sin A + \csc A)^2\)[/tex]:
[tex]\[ (\sin A + \csc A)^2 = \sin^2 A + 2 \sin A \csc A + \csc^2 A \][/tex]
Since [tex]\(\csc A = \frac{1}{\sin A}\)[/tex]:
[tex]\[ = \sin^2 A + 2 \sin A \cdot \frac{1}{\sin A} + \left( \frac{1}{\sin A} \right)^2 = \sin^2 A + 2 + \csc^2 A = \sin^2 A + 2 + \frac{1}{\sin^2 A} = 1 - \cos^2 A + 2 + \frac{1}{\sin^2 A} \][/tex]
2. For [tex]\((\cos A + \sec A)^2\)[/tex]:
[tex]\[ (\cos A + \sec A)^2 = \cos^2 A + 2 \cos A \sec A + \sec^2 A \][/tex]
Since [tex]\(\sec A = \frac{1}{\cos A}\)[/tex]:
[tex]\[ = \cos^2 A + 2 \cos A \cdot \frac{1}{\cos A} + \left( \frac{1}{\cos A} \right)^2 = \cos^2 A + 2 + \sec^2 A = \cos^2 A + 2 + \frac{1}{\cos^2 A} = 1 - \sin^2 A + 2 + \frac{1}{\cos^2 A} \][/tex]
Now combine the results:
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = (1 - \cos^2 A + 2 + \frac{1}{\sin^2 A}) + (1 - \sin^2 A + 2 + \frac{1}{\cos^2 A}) = 4 + (1 - \cos^2 A + \frac{1}{\sin^2 A}) + (1 - \sin^2 A + \frac{1}{\cos^2 A}) \][/tex]
Further simplification shows:
[tex]\[ (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = \frac{10 \cos^2 A - 18}{\cos(4A) - 1} \][/tex]
### Step 2: Simplify the Right-Hand Side (RHS)
The RHS is:
[tex]\[ \tan^2 A + \cot 2A + 7 \][/tex]
### Step 3: Check if LHS = RHS
To verify the equality, we can check if the simplified forms for LHS and RHS are equivalent or not:
Given:
[tex]\[ \text{LHS} = \frac{10 \cos(2A)^2 - 18}{\cos(4A) - 1} \][/tex]
[tex]\[ \text{RHS} = \tan^2 A + \cot 2A + 7 \][/tex]
### Step 4: Solution
We compare the simplified forms:
[tex]\[ \frac{10 \cos(2A)^2 - 18}{\cos(4A) - 1} \neq \tan^2 A + \cot 2A + 7 \][/tex]
Thus, the given equation does not hold true for all values of [tex]\(A\)[/tex]. Therefore, the solution to the given equation is that the LHS does not equal the RHS under general assumptions.
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