Find the value of other five trigonometric function: x = - 5 12 , x lies in second quadrant.

Here x = - 5 12 x = 1 x = - 12 5 Now ^2 x = 1 + ^2 x ^2x = 1 + ( - 5 12 )^2 ^2x = 169 144 x = 13 12 But x lies in second quadrant. sex = - 13 12 x = 1 x = - 12 13 Also ^2 x + ^2 x = 1 ^2x + ( - 12 13 )^2 = 1 ^2x = 1 - 144 169 ^2x = 25 169 x = 5 13 But x lies in second quadrant. x = 5 13 ecx = 1 x = 13 5

Find the value of other five trigonometric function: x = - 5 12 ,…

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Question

Find the value of other five trigonometric function: $\tan x = \frac{{ - 5}}{{12}}$, x lies in second quadrant.

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Answer

Here $\tan x = \frac{{ - 5}}{{12}}$
$\cot x = \frac{1}{{\tan x}} = \frac{{ - 12}}{5}$
Now ${\sec ^2}x = 1 + {\tan ^2}x \Rightarrow {\sec ^2}x = 1 + {\left( {\frac{{ - 5}}{{12}}} \right)^2}$$ \Rightarrow {\sec ^2}x = \frac{{169}}{{144}}$
$ \Rightarrow \sec x = \pm \frac{{13}}{{12}}$
But x lies in second quadrant.
$\therefore sex = \frac{{ - 13}}{{12}}$
$\cos x = \frac{1}{{\sec x}} = \frac{{ - 12}}{{13}}$
Also ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x + {\left( {\frac{{ - 12}}{{13}}} \right)^2} = 1$$ \Rightarrow {\sin ^2}x = 1 - \frac{{144}}{{169}}$
$ \Rightarrow {\sin ^2}x = \frac{{25}}{{169}} \Rightarrow \sin x = \pm \frac{5}{{13}}$
But x lies in second quadrant.
$\therefore \sin x = \frac{5}{{13}}$
$\cos ecx = \frac{1}{{\sin x}} = \frac{{13}}{5}$

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