QPaperGen

\(\textbf{Step 1: Determine quadrant of \(\frac{x}{2}\):}\)

Since \(x\) lies in the second quadrant, i.e., \(\frac{\pi}{2} < x < \pi\), then:
\[
\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}
\]
So, \(\frac{x}{2}\) lies in the first quadrant. Therefore:
\[
\sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2} \text{ are all positive}
\]

\(\textbf{Step 2: Use identity for \(\sec^2 x\)}\)
\[
\sec^2 x = 1 + \tan^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}
\]
Since \(x\) is in the second quadrant, \(\sec x < 0\), so:
\[
\sec x = -\frac{5}{3} \Rightarrow \cos x = -\frac{3}{5}
\]

\(\textbf{Step 3: Use half-angle identity for cosine}\)
\[
2\cos^2\left(\frac{x}{2}\right) = 1 + \cos x = 1 - \frac{3}{5} = \frac{2}{5}
\Rightarrow \cos^2\left(\frac{x}{2}\right) = \frac{1}{5}
\Rightarrow \cos\left(\frac{x}{2}\right) = \frac{1}{\sqrt{5}}
\]

\(\textbf{Step 4: Use Pythagorean identity}\)
\[
\sin^2\left(\frac{x}{2}\right) = 1 - \cos^2\left(\frac{x}{2}\right) = 1 - \frac{1}{5} = \frac{4}{5}
\Rightarrow \sin\left(\frac{x}{2}\right) = \frac{2}{\sqrt{5}}
\]

\(\textbf{Step 5: Compute tangent}\)
\[
\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2
\]


\(\textbf{Final Answer:}\)
\[
\sin\left(\frac{x}{2}\right) = \frac{2\sqrt{5}}{5}, \quad
\cos\left(\frac{x}{2}\right) = \frac{\sqrt{5}}{5}, \quad
\tan\left(\frac{x}{2}\right) = 2
\]