To find [tex]\(\tan \alpha\)[/tex] and [tex]\(\sec \alpha\)[/tex] given that [tex]\(\sin \alpha = \frac{2 \sqrt{a}}{a + 1}\)[/tex]:
1. Calculate [tex]\(\cos \alpha\)[/tex]:
Using the Pythagorean identity for sine and cosine, we have:
[tex]\[
\sin^2 \alpha + \cos^2 \alpha = 1
\][/tex]
Given [tex]\(\sin \alpha = \frac{2 \sqrt{a}}{a + 1}\)[/tex], we first determine [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[
\sin^2 \alpha = \left( \frac{2 \sqrt{a}}{a + 1} \right)^2 = \frac{4a}{(a + 1)^2}
\][/tex]
Substituting [tex]\(\sin^2 \alpha\)[/tex] into the Pythagorean identity:
[tex]\[
\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4a}{(a + 1)^2}
\][/tex]
Simplifying the expression:
[tex]\[
\cos^2 \alpha = \frac{(a + 1)^2 - 4a}{(a + 1)^2}
\][/tex]
[tex]\[
\cos^2 \alpha = \frac{a^2 + 2a + 1 - 4a}{(a + 1)^2} = \frac{a^2 - 2a + 1}{(a + 1)^2} = \frac{(a - 1)^2}{(a + 1)^2}
\][/tex]
Taking the square root of both sides gives us [tex]\(\cos \alpha\)[/tex]:
[tex]\[
\cos \alpha = \frac{|a - 1|}{a + 1}
\][/tex]
2. Calculate [tex]\(\tan \alpha\)[/tex]:
Using the definition [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex], we substitute the values we have found:
[tex]\[
\tan \alpha = \frac{\frac{2\sqrt{a}}{a+1}}{\frac{|a-1|}{a+1}} = \frac{2\sqrt{a}}{|a-1|}
\][/tex]
3. Calculate [tex]\(\sec \alpha\)[/tex]:
Using the definition [tex]\(\sec \alpha = \frac{1}{\cos \alpha}\)[/tex], we substitute the value of [tex]\(\cos \alpha\)[/tex]:
[tex]\[
\sec \alpha = \frac{1}{\frac{|a - 1|}{a + 1}} = \frac{a + 1}{|a - 1|}
\][/tex]
Hence, the values are:
[tex]\[
\tan \alpha = \frac{2 \sqrt{a}}{|a - 1|}
\][/tex]
[tex]\[
\sec \alpha = \frac{a + 1}{|a - 1|}
\][/tex]
