Here given that, \( tanx=-\dfrac{5}{12}\)
Therefore we have,
\( cotx=\dfrac{1}{tanx}\)
\( \dfrac{1}{(-\dfrac{5}{12})}= -\dfrac{12}{5}\)
We know that
sec2 x - tan2 x = 1
We can write it as
sec2 x = 1 + tan2 x
Substituting tanx = \(- \dfrac{5}{12}\) in the formula, we obtain,
\( sec^2x=1+(-\dfrac{5}{12})^2\)
\( sec^2x=1+\dfrac{25}{144}\)
= \( \dfrac{169}{144}\)
= \( secx\pm \dfrac{13}{12}\)
Since x lies in the 2nd quadrant, the value of secx will be negative.
\( secx=- \dfrac{13}{12}\)
Therefore, \( cosx=\dfrac{1}{secx}\)
= \(- \dfrac{12}{13}\)
\( tanx=\dfrac{sinx}{cosx}\)
Therefore, sinx=tanxcosx
Hence we have,\( sinx=(-\dfrac{5}{12})\times - \dfrac{12}{13}\)
\(=( \dfrac{5}{13})\)
\( cosecx=\dfrac{1}{sinx}= \dfrac{13}{5}\)
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