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Find the values of other five trigonometric functions cosx = -1\2,x lies in third quadrant

Q (1) Find the values of other five trigonometric functions

\small \cos x = -\frac{1}{2}  , x lies in third quadrant.

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Solution 
\cos x = -\frac {1}{2}
   \because \sec x = \frac{1}{\cos x} = \frac{1}{-\frac {1}{2}} = -2
x lies in III quadrants.  Therefore sec x is negative

\sin ^{2}x +\cos^{2}x = 1 \\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 -\left ( -\frac{1}{2} \right )^{2}\\ \sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}\\ \sin x = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}
    x lies in III quadrants.  Therefore sin x is negative
\therefore \sin x= - \frac{\sqrt{3}}{2}

\because cosec \ x = \frac {1}{\sin x}= \frac{1}{- \frac{\sqrt{3}}{2}} =- \frac{2}{\sqrt{3}}

x lies in III quadrants.  Therefore cosec x is negative

\tan x = \frac{\sin x}{\cos x} = \frac {-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}
x lies in III quadrants.  Therefore tan x is positive

\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}
x lies in III quadrants.  Therefore cot x is positive
 

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