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