Sin a=12/13 , P/2<a<p

Given that \(\sin a = \frac{12}{13}\) and the range is \(\frac{\pi}{2} < a < \pi\), the angle \(a\) lies in the second quadrant where the sine is positive and the cosine is negative.

To find \(\cos a\), we can use the Pythagorean identity:

\[ \sin^2 a + \cos^2 a = 1 \]

Substituting for \(\sin a\):

\[ \left( \frac{12}{13} \right)^2 + \cos^2 a = 1 \]

Calculating \(\left( \frac{12}{13} \right)^2\):

\[ \frac{144}{169} + \cos^2 a = 1 \]

Now, subtract \(\frac{144}{169}\) from both sides:

\[ \cos^2 a = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \]

Now, taking the square root of both sides gives us the cosine value:

\[ \cos a = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \]

Since \(a\) is in the second quadrant, \(\cos a\) is negative:

\[ \cos a = -\frac{5}{13} \]

Now, we have:

• \(\sin a = \frac{12}{13}\)
• \(\cos a = -\frac{5}{13}\)

These values can be used for further calculations, such as finding \(\tan a\), \(\sec a\), \(\csc a\), and \(\cot a\) if needed.