It is given that
sin x = \( \dfrac{3}{5}\)
We can write it as
\( cosecx=\dfrac{1}{sinx}\)
= \(\dfrac{1}{\dfrac{3}{5}}\)= \( \dfrac{5}{3}\)
We know that
sin2 x + cos2 x = 1
We can write it as
cos2 x = 1 – sin2 x
Substituting sin x =\( \dfrac{3}{5}\) in the formula, we obtain,
\( cos^2x=1- (\dfrac{3}{5})^2\)
\( cos^2x=1- (\dfrac{9}{25})\) = \( \dfrac{16}{25}\)
cosx= \( \pm \dfrac{4}{5}\)
Since x lies in the 2nd quadrant, the value of cos x will be negative
cosx = \(-\dfrac{4}{5}\)
Therefore, secx = \( \dfrac{1}{cosx}\)
= \( \dfrac{1}{(-\dfrac{4}{5})}=- \dfrac{5}{4}\)
Hence,
\( tanx=\dfrac{sinx}{cosx}\)
= \(\dfrac{(\dfrac{3}{5})}{(-\dfrac{4}{5})}=\dfrac{1}{2}=-\dfrac{3}{4}\)
And
\( cotx=\dfrac{1}{tanx}=-\dfrac{4}{3}\)
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