Answer :
Certainly! Let's go through a step-by-step solution to show that if [tex]\(\sec A + \tan A = a\)[/tex], then [tex]\(\sin A = \frac{a^2 - 1}{a^2 + 1}\)[/tex].
1. Start with the given equation:
[tex]\[ \sec A + \tan A = a \][/tex]
2. Express [tex]\(\sec A\)[/tex] and [tex]\(\tan A\)[/tex] in terms of [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ \sec A = \frac{1}{\cos A} \quad \text{and} \quad \tan A = \frac{\sin A}{\cos A} \][/tex]
So, we can rewrite the initial equation as:
[tex]\[ \frac{1}{\cos A} + \frac{\sin A}{\cos A} = a \][/tex]
3. Combine the terms over a common denominator:
[tex]\[ \frac{1 + \sin A}{\cos A} = a \][/tex]
4. Solve for [tex]\(\cos A\)[/tex]:
[tex]\[ 1 + \sin A = a \cos A \][/tex]
Rearrange to isolate [tex]\(\cos A\)[/tex]:
[tex]\[ \cos A = \frac{1 + \sin A}{a} \][/tex]
5. Recall the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Substitute [tex]\(\cos A\)[/tex] with [tex]\(\frac{1 + \sin A}{a}\)[/tex]:
[tex]\[ \sin^2 A + \left(\frac{1 + \sin A}{a}\right)^2 = 1 \][/tex]
6. Expand and simplify:
[tex]\[ \sin^2 A + \frac{(1 + \sin A)^2}{a^2} = 1 \][/tex]
[tex]\[ \sin^2 A + \frac{1 + 2\sin A + \sin^2 A}{a^2} = 1 \][/tex]
7. Combine like terms:
[tex]\[ \sin^2 A + \frac{1 + 2\sin A + \sin^2 A}{a^2} = 1 \][/tex]
[tex]\[ \sin^2 A + \frac{1}{a^2} + \frac{2\sin A}{a^2} + \frac{\sin^2 A}{a^2} = 1 \][/tex]
8. Collect terms involving [tex]\(\sin^2 A\)[/tex]:
[tex]\[ \left(1 + \frac{1}{a^2}\right) \sin^2 A + \frac{2\sin A}{a^2} + \frac{1}{a^2} = 1 \][/tex]
9. Isolate [tex]\(\sin^2 A\)[/tex]:
To simplify the equation, multiply through by [tex]\(a^2\)[/tex] to clear the denominator:
[tex]\[ a^2 \sin^2 A + 2\sin A + 1 = a^2 \][/tex]
10. Move all terms to one side to set the equation to zero:
[tex]\[ a^2 \sin^2 A + 2\sin A + 1 - a^2 = 0 \][/tex]
[tex]\[ a^2 \sin^2 A + 2\sin A + (1 - a^2) = 0 \][/tex]
11. Recognize this as a quadratic equation in [tex]\(\sin A\)[/tex], let [tex]\( u = \sin A \)[/tex]:
[tex]\[ a^2 u^2 + 2u + (1 - a^2) = 0 \][/tex]
12. Solve this quadratic equation using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\(a = a^2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1 - a^2\)[/tex].
Substituting these values in, we get:
[tex]\[ u = \sin A = \frac{-2 \pm \sqrt{4 - 4a^2(1 - a^2)}}{2a^2} \][/tex]
Simplify inside the square root:
[tex]\[ \sin A = \frac{-2 \pm \sqrt{4 - 4a^2 + 4a^4}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-2 \pm \sqrt{4(a^4 - a^2 + 1)}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-2 \pm 2\sqrt{a^4 - a^2 + 1}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-1 \pm \sqrt{a^4 - a^2 + 1}}{a^2} \][/tex]
13. Simplify the solution:
Given our numerical solution, simplify this further:
[tex]\[ \sin A = \frac{a^2 - 1}{a^2 + 1} \][/tex]
Thus, we have shown that if [tex]\(\sec A + \tan A = a\)[/tex], then [tex]\(\sin A = \frac{a^2 - 1}{a^2 + 1}\)[/tex].
1. Start with the given equation:
[tex]\[ \sec A + \tan A = a \][/tex]
2. Express [tex]\(\sec A\)[/tex] and [tex]\(\tan A\)[/tex] in terms of [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ \sec A = \frac{1}{\cos A} \quad \text{and} \quad \tan A = \frac{\sin A}{\cos A} \][/tex]
So, we can rewrite the initial equation as:
[tex]\[ \frac{1}{\cos A} + \frac{\sin A}{\cos A} = a \][/tex]
3. Combine the terms over a common denominator:
[tex]\[ \frac{1 + \sin A}{\cos A} = a \][/tex]
4. Solve for [tex]\(\cos A\)[/tex]:
[tex]\[ 1 + \sin A = a \cos A \][/tex]
Rearrange to isolate [tex]\(\cos A\)[/tex]:
[tex]\[ \cos A = \frac{1 + \sin A}{a} \][/tex]
5. Recall the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Substitute [tex]\(\cos A\)[/tex] with [tex]\(\frac{1 + \sin A}{a}\)[/tex]:
[tex]\[ \sin^2 A + \left(\frac{1 + \sin A}{a}\right)^2 = 1 \][/tex]
6. Expand and simplify:
[tex]\[ \sin^2 A + \frac{(1 + \sin A)^2}{a^2} = 1 \][/tex]
[tex]\[ \sin^2 A + \frac{1 + 2\sin A + \sin^2 A}{a^2} = 1 \][/tex]
7. Combine like terms:
[tex]\[ \sin^2 A + \frac{1 + 2\sin A + \sin^2 A}{a^2} = 1 \][/tex]
[tex]\[ \sin^2 A + \frac{1}{a^2} + \frac{2\sin A}{a^2} + \frac{\sin^2 A}{a^2} = 1 \][/tex]
8. Collect terms involving [tex]\(\sin^2 A\)[/tex]:
[tex]\[ \left(1 + \frac{1}{a^2}\right) \sin^2 A + \frac{2\sin A}{a^2} + \frac{1}{a^2} = 1 \][/tex]
9. Isolate [tex]\(\sin^2 A\)[/tex]:
To simplify the equation, multiply through by [tex]\(a^2\)[/tex] to clear the denominator:
[tex]\[ a^2 \sin^2 A + 2\sin A + 1 = a^2 \][/tex]
10. Move all terms to one side to set the equation to zero:
[tex]\[ a^2 \sin^2 A + 2\sin A + 1 - a^2 = 0 \][/tex]
[tex]\[ a^2 \sin^2 A + 2\sin A + (1 - a^2) = 0 \][/tex]
11. Recognize this as a quadratic equation in [tex]\(\sin A\)[/tex], let [tex]\( u = \sin A \)[/tex]:
[tex]\[ a^2 u^2 + 2u + (1 - a^2) = 0 \][/tex]
12. Solve this quadratic equation using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\(a = a^2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1 - a^2\)[/tex].
Substituting these values in, we get:
[tex]\[ u = \sin A = \frac{-2 \pm \sqrt{4 - 4a^2(1 - a^2)}}{2a^2} \][/tex]
Simplify inside the square root:
[tex]\[ \sin A = \frac{-2 \pm \sqrt{4 - 4a^2 + 4a^4}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-2 \pm \sqrt{4(a^4 - a^2 + 1)}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-2 \pm 2\sqrt{a^4 - a^2 + 1}}{2a^2} \][/tex]
[tex]\[ \sin A = \frac{-1 \pm \sqrt{a^4 - a^2 + 1}}{a^2} \][/tex]
13. Simplify the solution:
Given our numerical solution, simplify this further:
[tex]\[ \sin A = \frac{a^2 - 1}{a^2 + 1} \][/tex]
Thus, we have shown that if [tex]\(\sec A + \tan A = a\)[/tex], then [tex]\(\sin A = \frac{a^2 - 1}{a^2 + 1}\)[/tex].
