Q (2) Find the values of other five trigonometric functions

\small \sin x = \frac{3}{5}  x lies in second quadrant.

Answers (1)
S safeer

Solution 

\sin x = \frac {3}{5}

cosec \ x = \frac{1}{\sin x}=\frac {1}{\frac {3}{5}} = \frac {5}{3}
x lies in the second quadrant.  Therefore cosec x is positive

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

\sec x = \frac {1}{\cos x} = \frac{1}{- \frac {4}{5}} = -\frac {5}{4}
x lies in the second quadrant.  Therefore sec x is negative

 \tan x = \frac {\sin x}{\cos x} = \frac {\frac{3}{5}}{-\frac{4}{5}} = -\frac {3}{4}
x lies in the second quadrant.  Therefore tan x is negative

\cot x = \frac {1}{\tan x} = \frac {1}{-\frac {3}{4}} = -\frac{4}{3}
x lies in the second quadrant.  Therefore cot x is negative

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