Find the exact value of $ an( heta) $ given that $ sin( heta) = frac{3}{5} $ and $ heta $ is in the second quadrant.

Given that $ \sin(\theta) = \frac{3}{5} $ and $ \theta $ is in the second quadrant:

Since $ \sin(\theta) $ is positive in the second quadrant, $ \cos(\theta) $ must be negative:

Use the Pythagorean identity:

$ \sin^2(\theta) + \cos^2(\theta) = 1 $

Substitute $ \sin(\theta) = \frac{3}{5} $:

$ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 $

$ \frac{9}{25} + \cos^2(\theta) = 1 $

$ \cos^2(\theta) = 1 – \frac{9}{25} = \frac{16}{25} $

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

$ \cos(\theta) = -\frac{4}{5} $

Now find $ \tan(\theta) $:

$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} $

Thus, $ \tan(\theta) = -\frac{3}{4} $.

Given $ sin( heta) = frac{3}{5} $ and $ heta $ in the second quadrant,

Using $ sin^2( heta) + cos^2( heta) = 1 $:

$ frac{9}{25} + cos^2( heta) = 1 $

$ cos^2( heta) = frac{16}{25} $

Since $ heta $ is in the second quadrant, $ cos( heta) = -frac{4}{5} $

$ an( heta) = frac{sin( heta)}{cos( heta)} = frac{frac{3}{5}}{-frac{4}{5}} = -frac{3}{4} $

Given $ sin( heta) = frac{3}{5} $ and $ heta $ is in the second quadrant,

$ cos( heta) = -frac{4}{5} $

$ an( heta) = frac{sin( heta)}{cos( heta)} = -frac{3}{4} $

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