Certainly! Let's go through the steps to find [tex]\(\tan \theta\)[/tex] given that [tex]\(\sin \theta = \frac{a-1}{a+1}\)[/tex].
First, we recognize that:
[tex]\[
\sin \theta = \frac{a-1}{a+1}
\][/tex]
Next, we use the Pythagorean identity for sine and cosine. We know that:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
From the given [tex]\(\sin \theta\)[/tex], we calculate [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[
\sin^2 \theta = \left(\frac{a-1}{a+1}\right)^2 = \frac{(a-1)^2}{(a+1)^2}
\][/tex]
By the Pythagorean identity, we can express [tex]\(\cos^2 \theta\)[/tex] as:
[tex]\[
\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{(a-1)^2}{(a+1)^2}
\][/tex]
To simplify this, we find a common denominator:
[tex]\[
\cos^2 \theta = \frac{(a+1)^2 - (a-1)^2}{(a+1)^2}
\][/tex]
Expand both squares:
[tex]\[
(a+1)^2 = a^2 + 2a + 1
\][/tex]
[tex]\[
(a-1)^2 = a^2 - 2a + 1
\][/tex]
Subtract these to get:
[tex]\[
(a+1)^2 - (a-1)^2 = (a^2 + 2a + 1) - (a^2 - 2a + 1) = 4a
\][/tex]
Thus:
[tex]\[
\cos^2 \theta = \frac{4a}{(a+1)^2}
\][/tex]
Taking the positive square root (considering that [tex]\(\theta\)[/tex] is in the quadrant where cosine is non-negative):
[tex]\[
\cos \theta = \sqrt{\frac{4a}{(a+1)^2}}
\][/tex]
This can be simplified further:
[tex]\[
\cos \theta = \frac{2\sqrt{a}}{a+1}
\][/tex]
Now, we can find [tex]\(\tan \theta\)[/tex] using the relationship:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Substitute the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\tan \theta = \frac{\frac{a-1}{a+1}}{\frac{2\sqrt{a}}{a+1}} = \frac{a-1}{2\sqrt{a}}
\][/tex]
Simplify the expression:
[tex]\[
\tan \theta = \frac{a-1}{(a+1) \cdot \frac{2\sqrt{a}}{a+1}} = \frac{a-1}{2\sqrt{a}}
\][/tex]
Therefore, the value of [tex]\(\tan \theta\)[/tex] is:
[tex]\[
\boxed{\tan \theta = \frac{a-1}{2\sqrt{a}}}
\][/tex]
So, the detailed step-by-step solution yields:
- [tex]\(\sin^2 \theta = \frac{(a-1)^2}{(a+1)^2}\)[/tex]
- [tex]\(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{\frac{4a}{(a+1)^2}} = \frac{2\sqrt{a}}{a+1}\)[/tex]
- [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{a-1}{2\sqrt{a}}\)[/tex]
This matches our derived result for [tex]\(\tan \theta\)[/tex].
