IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
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.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.
