It is given that
cot x =\( \dfrac{3}{4}\)
We can write it as
\(tanx= \dfrac{1}{cotx}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
We know that
1 + tan2 x = sec2 x
We can write it as
1 + \(( \dfrac{4}{3})^2\)= sec2 x
Substituting the values
1 + \( \dfrac{16}{9}\) = sec2 x
cos2 x = \( \dfrac{25}{9}\)
sec x = ± \( \dfrac{5}{3}\)
Here x lies in the third quadrant so the value of sec x will be negative
sec x = \(- \dfrac{5}{3}\)
We can write it as
Therefore,
\( cos x=\dfrac{1}{secx}\)
= \(\dfrac{1}{(-\dfrac{5}{3})}\)
= \(-\dfrac{3}{5}\)
Now, \( tanx=\dfrac{sinx}{cosx}\)
Therefore, sinx=tanxcosx
Hence we have, sinx = \( \dfrac{4}{3}\times (-\dfrac{3}{5})\)
= \((- \dfrac{4}{5})\)
And
\( cosecx=\dfrac{1}{sinx}\)
= \(- \dfrac{5}{4}\)
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