Find the values of other five trigonometric functions if cos x = $$- \dfrac{1}{2}$$, x lies in third quadrant.

Asked by Pragya Singh | 1 year ago |  67

##### Solution :-

Here given that, $$cosx=-\dfrac{1}{2}$$

Therefore we have,

$$secx=\dfrac{1}{cosx}$$

$$\dfrac{1}{(-\dfrac{1}{2})}$$ = -2

We know that

sin2 x + cos2 x = 1

We can write it as

sin2 x = 1 – cos2 x

Substituting cosx = $$- \dfrac{1}{2}$$ in the formula, we obtain,

$$sin^2x=1-(-\dfrac{1}{2})^2$$

$$sin^2x=1-\dfrac{1}{4}$$ = $$\dfrac{3}{4}$$

$$sinx\pm \dfrac{\sqrt{3}}{2}$$

Since x lies in the 3rd quadrant, the value of sin x will be negative.

$$sinx=- \dfrac{\sqrt{3}}{2}$$

Therefore, $$cosecx=\dfrac{1}{sinx}$$

$$\dfrac{1}{(\dfrac{-\sqrt{3}}{2})}=-\dfrac{2}{\sqrt{3}}$$

Hence,

$$tanx=\dfrac{sinx}{cosx}$$

$$\dfrac{(-\dfrac{\sqrt{3})}{2}}{(-\dfrac{1}{2})}=\sqrt{3}$$

And

$$cotx=\dfrac{1}{tanx}$$

$$\dfrac{1}{\sqrt{3}}$$

Answered by Abhisek | 1 year ago

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