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

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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