Answer:
\(cosec\theta\)=-\(\frac{5}{4}\)
\(tan\theta\)=\(\frac{4}{3}\).
Step-by-step explanation:
Given, let \(\theta\) be the angle in the III quadrant .
\(cos\theta\)=-\(\frac{3}{5}\)...... (given)
In III quadrant \(cosec\theta\) is negative and \(tan\theta\) is positive .
because , \(cosec\theta\)=reciprocal of \(sin\theta\)
it means , \(cosec\theta\)=\(\frac{1}{sin\theta}\)
Therefore, first we find value of \(sin\theta\)
We know that
\(sin^2\theta=1-cos^2\theta\)
∴ \(sin\theta=\sqrt{1-cos^2\theta}\)
\(sin\theta=\sqrt{1-(\frac{-3}{5} )^2\)
\(sin\theta=\sqrt{1-\frac{9}{25}}\)
\(sin\theta=\sqrt{\frac{16} {25}\)
\(sin\theta=\frac{-4}{5}\)
Because \(sin\theta\) lies in III quadrant and in III quadrant it is negative.
Now, we know that
\(tan\theta=\sqrt{sec^2\theta-1}\)
therefore, first we find \(sec\theta\)
\(sec\theta=\frac{1}{cos\theta}\)
\(sec\theta=-\frac{5}{3}\)
\(tan\theta=\sqrt{(\frac{5}{3})^2-1\)
\(tan\theta=\sqrt{\frac{25} {9}-1\)
\(tan\theta=\frac{4}{3}\)
Because it lies in III quadrant, therefore it take positive.
Hence,\(cosec\theta=-\frac{5}{4}\) and \(tan\theta=\frac{4}{3}\)
