Find the value of other five trigonometric function: x = 3 4 , x lies in third quadrant.

Here x = 3 4 x = 1 x = 4 3 Now ^2 x = 1 + ^2 x ^2x = 1 + ( 4 3 )^2 ^2x = 1 + 16 9 ^2x = 25 9 x = 5 3 But x lies in third quadrant. x = - 5 3 x = 1 x = - 3 5 Also ^2 x + ^2 x = 1 ^2x + ( - 3 5 )^2 = 1 ^2x = 1 - 9 25 ^2x = 16 25 x = 4 5 But x lies in third quadrant. x ;= - 4 5 ecx = 1 x = - 5 4

Find the value of other five trigonometric function: x = 3 4 , x …

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Question

Find the value of other five trigonometric function: $\cot x = \frac{3}{4}$, x lies in third quadrant.

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Answer

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

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