Question

if sin theta = 4/5 and is in quadrant 2, what is the value of tan theta

Answer 1

\[\huge \sin \theta = \frac{4}{5}\]
we know that:
\[\huge \sin^2 \theta + \cos^2 \theta = 1\]
\[\huge \rightarrow \cos^2 \theta = 1-\sin^2 \theta= 1-(\frac{4}{5})^2\]
\[\huge \rightarrow \cos^2 \theta = 1-\frac{16}{25}= \frac{25-16}{25}= \frac{9}{25}\]
\[\huge \rightarrow \cos^2 \theta = \frac{9}{25}\]
\[\huge \rightarrow \cos \theta = \pm \sqrt{ \frac{9}{25}}\]
\[\huge \rightarrow \cos \theta = \pm \frac{3}{5}\]
Since theta lies in the second quadrant, hence
\[\huge \rightarrow \cos \theta = - \frac{3}{5}\]
Now we know that:
\[\huge \tan \theta = \frac{\sin \theta }{ \cos \theta}\]
\[\huge \rightarrow \tan \theta = \frac{\frac{4}{5} }{\frac{-3}{5}}\]
\[\huge \rightarrow \tan \theta = \frac{4}{5} \times \frac{5}{-3}\]
\[\huge \rightarrow \tan \theta =\frac{4}{-3}\]
\[\huge \rightarrow \tan \theta =-\frac{4}{3}\]

@TSV