Find the value of other five trigonometric function: x = - 1 2 , x lies in third quadrant.

Here x = - 1 2 Now ^2 x + ^2 x = 1 ^2x + ( - 1 2 )^2 = 1 ^2x = 1 - 1 4 ^2x = 3 4 x = 3 2 But x lies in third quadrant. x = - 3 2 Now cosec x = 1 x = - 2 3 and x = 1 x = - 2 x = x = - 3 /2 - 1/2 = 3 and x = x x = - 1/2 - 3 /2 = 1 3

Find the value of other five trigonometric function: x = - 1 2 , …

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Find the value of other five trigonometric function: $\cos x = - \frac{1}{2}$, x lies in third quadrant.

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Answer

Here $\cos x = - \frac{1}{2}$
Now ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x + {\left( { - \frac{1}{2}} \right)^2} = 1$$ \Rightarrow {\sin ^2}x = 1 - \frac{1}{4}$
$ \Rightarrow {\sin ^2}x = \frac{3}{4} \Rightarrow \sin x = \pm \frac{{\sqrt 3 }}{2}$
But x lies in third quadrant.
$\therefore \sin x = - \frac{{\sqrt 3 }}{2}$
Now cosec x $ = \frac{1}{{\sin x}} = - \frac{2}{{\sqrt 3 }}$ and $\sec x = \frac{1}{{\cos x}} = - 2$
$\tan x = \frac{{\sin x}}{{\cos }} = \frac{{ - \sqrt 3 /2}}{{ - 1/2}} = \sqrt 3 $ and $\cot x = \frac{{\cos x}}{{\sin x}} = \frac{{ - 1/2}}{{ - \sqrt 3 /2}} = \frac{1}{{\sqrt 3 }}$

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