Find the values of other five trigonometric functions if \( tan x=-\dfrac{5}{12}\), x lies in second quadrant.

Asked by Pragya Singh | 2 years ago |  86

1 Answer

Solution :-

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,\( sinx=(-\dfrac{5}{12})\times - \dfrac{12}{13}\)

\(=( \dfrac{5}{13})\)

\( cosecx=\dfrac{1}{sinx}= \dfrac{13}{5}\)

Answered by Abhisek | 2 years ago

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