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To solve the given expression [tex]\((\sec A + \tan A - 1) \cdot (\sec A - \tan A + 1) = 2 \tan A\)[/tex], let's proceed through a step-by-step solution.
1. Identify Relevant Trigonometric Identities:
- We know that [tex]\(\sec^2 A = 1 + \tan^2 A\)[/tex].
- Let us substitute the identities to simplify the expression. Denote [tex]\(\sec A\)[/tex] by [tex]\(x\)[/tex] and [tex]\(\tan A\)[/tex] by [tex]\(y\)[/tex]. So, we have:
[tex]\[ x = \sec A \quad \text{and} \quad y = \tan A \][/tex]
2. Rewrite the Given Expression:
- Substitute [tex]\(x\)[/tex] and [tex]\(y\)[/tex] into the expression:
[tex]\[ (x + y - 1) \cdot (x - y + 1) \][/tex]
3. Simplify the Expression:
- Expand the product:
[tex]\[ (x + y - 1)(x - y + 1) \][/tex]
Using the distributive property of multiplication over addition, this becomes:
[tex]\[ (x + y - 1) \cdot (x - y + 1) = \left( x(x - y + 1) \right) + \left( y(x - y + 1) \right) - \left( 1(x - y + 1) \right) \][/tex]
4. Further Expansion:
- Perform the multiplication step-by-step:
[tex]\[ x(x - y + 1) = x^2 - xy + x \][/tex]
[tex]\[ y(x - y + 1) = xy - y^2 + y \][/tex]
[tex]\[ -1(x - y + 1) = -x + y - 1 \][/tex]
- Combining these, we get:
[tex]\[ x^2 - xy + x + xy - y^2 + y - x + y - 1 \][/tex]
5. Combine Like Terms:
- Combine the like terms:
[tex]\[ x^2 - y^2 + y + y - 1 - \quad (xy \text{ and } -xy \text{ cancel out }\text{as these terms are} \text{zero}) \][/tex]
- We then have:
[tex]\[ x^2 - y^2 + 2y - 1 \][/tex]
6. Substitute [tex]\(\sec^2 A\)[/tex] back:
- Recall from the identity [tex]\( \sec^2 A = 1 + \tan^2 A \)[/tex], thus [tex]\( x^2 = 1 + y^2 \)[/tex]:
\\
[tex]\[ (1 + y^2) - y^2 + 2y - 1 \][/tex]
7. Simplification:
- Perform the simplification:
[tex]\[ 1 + y^2 - y^2 + 2y - 1 = 2y \][/tex]
- This results in:
[tex]\[ 2y \][/tex]
Therefore,
[tex]\[ (\sec A + \tan A - 1) \cdot (\sec A - \tan A + 1) = 2 \tan A \][/tex]
This confirms that the original expression is indeed equal to [tex]\(2 \tan A\)[/tex].
1. Identify Relevant Trigonometric Identities:
- We know that [tex]\(\sec^2 A = 1 + \tan^2 A\)[/tex].
- Let us substitute the identities to simplify the expression. Denote [tex]\(\sec A\)[/tex] by [tex]\(x\)[/tex] and [tex]\(\tan A\)[/tex] by [tex]\(y\)[/tex]. So, we have:
[tex]\[ x = \sec A \quad \text{and} \quad y = \tan A \][/tex]
2. Rewrite the Given Expression:
- Substitute [tex]\(x\)[/tex] and [tex]\(y\)[/tex] into the expression:
[tex]\[ (x + y - 1) \cdot (x - y + 1) \][/tex]
3. Simplify the Expression:
- Expand the product:
[tex]\[ (x + y - 1)(x - y + 1) \][/tex]
Using the distributive property of multiplication over addition, this becomes:
[tex]\[ (x + y - 1) \cdot (x - y + 1) = \left( x(x - y + 1) \right) + \left( y(x - y + 1) \right) - \left( 1(x - y + 1) \right) \][/tex]
4. Further Expansion:
- Perform the multiplication step-by-step:
[tex]\[ x(x - y + 1) = x^2 - xy + x \][/tex]
[tex]\[ y(x - y + 1) = xy - y^2 + y \][/tex]
[tex]\[ -1(x - y + 1) = -x + y - 1 \][/tex]
- Combining these, we get:
[tex]\[ x^2 - xy + x + xy - y^2 + y - x + y - 1 \][/tex]
5. Combine Like Terms:
- Combine the like terms:
[tex]\[ x^2 - y^2 + y + y - 1 - \quad (xy \text{ and } -xy \text{ cancel out }\text{as these terms are} \text{zero}) \][/tex]
- We then have:
[tex]\[ x^2 - y^2 + 2y - 1 \][/tex]
6. Substitute [tex]\(\sec^2 A\)[/tex] back:
- Recall from the identity [tex]\( \sec^2 A = 1 + \tan^2 A \)[/tex], thus [tex]\( x^2 = 1 + y^2 \)[/tex]:
\\
[tex]\[ (1 + y^2) - y^2 + 2y - 1 \][/tex]
7. Simplification:
- Perform the simplification:
[tex]\[ 1 + y^2 - y^2 + 2y - 1 = 2y \][/tex]
- This results in:
[tex]\[ 2y \][/tex]
Therefore,
[tex]\[ (\sec A + \tan A - 1) \cdot (\sec A - \tan A + 1) = 2 \tan A \][/tex]
This confirms that the original expression is indeed equal to [tex]\(2 \tan A\)[/tex].
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