Question

Find the other trigonometric functions if If cot x=3/4,xbox lies in the third quadrant​

Answer 1

Given :-

$\sf cotx = \dfrac{3}{4} \ (Lies \ in \ 3^{rd} \ quadrant )$  

To Find :-

$\bullet \sf \ tanx \\\\\bullet \ cosecx \\\\\bullet \ sinx \\\\\bullet \ cosx \\\\\bullet \ secx$

Solution :-

We know :-

$\bf tanx = \dfrac{1}{cotx}$

$\therefore \sf tanx = \dfrac{4}{3}$

We know :-

$\bf cosec^2x = 1+cot^2x$

$\longrightarrow \sf cosec^2x= 1+\dfrac{3}{4}^2 \\\\\longrightarrow cosec^2x= 1+\dfrac{9}{16} \\\\\longrightarrow cosec^2x = \dfrac{16+9}{16} \\\\\longrightarrow cosec^2x = \dfrac{25}{16} \\\\\longrightarrow cosecx = \sqrt{\dfrac{25}{16}} \\\\\longrightarrow cosecx = \pm \dfrac{5}{4}$

Since x lies in the third quadrant, cosec x will be negative.

$\therefore \sf cosecx = - \dfrac{5}{4}$

We know :-

$\bf sinx = \dfrac{1}{cosecx}$

$\therefore \sf sinx = -\dfrac{4}{5}$

We know :-

$\bf cosx = cotx \times sinx$

$\therefore \sf cosx = \dfrac{3}{4} \times -\dfrac{4}{5}$

          $= \sf - \dfrac{3}{5}$

We know :-

$\bf secx = \dfrac{1}{cosx}$

$\therefore \sf secx = - \dfrac{5}{3}$

Answer 2

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$\\ \rightarrow \sf{Cot \: x = \frac{3}{4} where \: x \: lies \: in \: 3rd \: quadrant.}$

$\rightarrow\sf{In \: 3rd \: quadrant \: :- \: tan \: x, cot \: x \: }\\\\\sf{ are \: positive \: . all \: rest \: are \: negative.}$

$\\\rightarrow \sf{Cot \: x = \frac{3}{4} = \frac{b}{p} }$

$\rightarrow \sf{h = \sqrt{(b {}^{2b} +P {}^{2} )} = \sqrt{(3 {}^{2} +4 {}^{2} )} = + \: or \: - 5} \\ \\ \rightarrow\sf{=5 tan \: x = \frac{1}{cot \: x } = \frac{4}{3} }$

$\\ \rightarrow \sf{Cos \: x = \frac{b}{h} = \frac{ - 3}{5} }\\ \\ \rightarrow\sf{ Sec \: x = \frac{1}{cos \: x} = \frac{ - 5}{3} }$

$\\ \rightarrow \sf{sin \: x = \frac{p}{h} = \frac{ -4}{5} }$

$\\ \rightarrow \sf{Cosec \: x = \frac{ -5}{4} }$

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$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

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