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Q (3) Find the values of other five trigonometric functions

\small \cot x = \frac{3}{4}  , x lies in third quadrant.

Answers (1)

Solution 

\cot x= \frac {3}{4}

\tan x = \frac{1}{\cot x}= \frac{1}{\frac {3}{4}} = \frac {4}{3}
1 + \tan ^ {2}x = \sec ^{2}x\\ 1+\frac{4^2}{3^2} = \sec ^{2}x\\ \\ 1 + \frac {16}{9} = \sec ^{2}x\\ \frac {25}{9} = \sec ^{2}x\\ \sec x = \sqrt {\frac {25}{9}} = \pm \frac {5}{3}
x lies in x lies in  third quadrant. therefore sec x is negative 
\sec x = -\frac{5}{3}

\cos x = \frac {1}{\sec x} = \frac {1}{-\frac{5}{3}} = -\frac{3}{5}
\sin ^{2 }x+ \cos ^{2}x = 1\\ \sin ^{2 }x = 1 - \cos ^{2}x\\ \sin ^{2 }x = 1 -\left ( -\frac{3}{5} \right )^{2}\\ \sin ^{2 }x = 1 - \frac {9}{25}\\ \sin ^{2 }x = \frac{16}{25}\\ \sin x = \sqrt {\frac{16}{25}} = \pm \frac{4}{5}
x lies in  x lies in  third quadrant. Therefore sin x is negative 
\sin x = -\frac {4}{5}
cosec x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}

Posted by

Safeer PP

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