Find the value of other five trigonometric function: sin x = 3 5 , x lies in second quadrant.

Here x = 3 5 Now ^2 x + ^2 x = 1 ( 3 5 )^2 + ^2 x = 1 ^2x = 1 - 9 25 ^2x = 16 25 x = 4 5 But x lies in second quadrant. x = - 4 5 Now ec ;x = 1 x = 5 3 and ;x = 1 x = - 5 4 x = x x = 3/5 - 4/5 = - 3 4 and x = x x = - 4/5 3/5 = - 4 3

Find the value of other five trigonometric function: sin x = 3 5 …

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Question

Find the value of other five trigonometric function: sin $x = \frac{3}{5}$, x lies in second quadrant.

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Answer

Here $\sin x = \frac{3}{5}$
Now ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\left( {\frac{3}{5}} \right)^2} + {\cos ^2}x = 1$$ \Rightarrow {\cos ^2}x = 1 - \frac{9}{{25}}$
$ \Rightarrow {\cos ^2}x = \frac{{16}}{{25}} \Rightarrow \cos x = \pm \frac{4}{5}$
But x lies in second quadrant.
$\therefore \cos x = - \frac{4}{5}$
Now $\cos ec\;x = \frac{1}{{\sin x}} = \frac{5}{3}$ and $\sec \;x = \frac{1}{{\cos x}} = - \frac{5}{4}$
$\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{3/5}}{{ - 4/5}} = \frac{{ - 3}}{4}$ and $\cot x = \frac{{\cos x}}{{\sin x}} = \frac{{ - 4/5}}{{3/5}} = \frac{{ - 4}}{3}$

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