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

Asked by Pragya Singh | 2 years ago |  87

1 Answer

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 | 2 years ago

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