Question

$\huge \mathcal \red{Question:}$

$\sf If \: cos \: x = \frac{ - 3}{5} \: \\ \sf \: and \: π<x< \frac{3\pi}{2} \sf \: \\ \sf \: find \: \: the \: values \: \\ \sf \: of \: other \: five \: trigonometric \\ \sf \: functions \: and \: hence \: evaluate. \\ \frac{cosec \: x + \: cot \: x}{sec \: x \: - tan \: x}$
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Answer 1

$\large\underline{\bold{Solution-}}$

$\sf \: \: cos \: x = \dfrac{ - 3}{5} \: and \: \pi \: < x < \dfrac{3\pi}{2}$

We know,

$\rm :\longmapsto\: {sin}^{2} x + {cos}^{2}x = 1$

$\rm :\longmapsto\: {sin}^{2} x + \dfrac{9}{25} = 1$

$\rm :\longmapsto\: {sin}^{2} x = 1 - \dfrac{9}{25}$

$\rm :\longmapsto\: {sin}^{2} x = \dfrac{25 - 9}{25}$

$\rm :\longmapsto\: {sin}^{2} x = \dfrac{16}{25}$

$\rm :\implies\:sinx \: = \: \pm \: \dfrac{4}{5}$

As

$\bf \: \pi \: < x < \dfrac{3\pi}{2} \: i.e. \: x \: lies \: in \: {3}^{rd} \: quadrant$

$\rm :\implies\:sinx \: < \: 0$

$\bf\implies \:sinx \: = \: - \: \dfrac{4}{5}$

Now,

we know that

$\rm :\longmapsto\:tanx \: = \: \dfrac{sinx}{cosx}$

$\bf\implies \:tanx = \dfrac{ - 4}{5} \div \dfrac{ - 3}{5} = \dfrac{4}{3}$

Now,

$\rm :\longmapsto\:secx = \dfrac{1}{cosx} = - \dfrac{5}{3}$

$\rm :\longmapsto\:cosecx = \dfrac{1}{sinx} = - \dfrac{5}{4}$

$\rm :\longmapsto\:cotx = \dfrac{1}{tanx} = \dfrac{3}{4}$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{cosec \: x + \: cot \: x}{sec \: x \: - tan \: x}$

$\sf \: = \: \dfrac{ - \dfrac{5}{4} + \dfrac{3}{4} }{ - \dfrac{5}{3} - \dfrac{4}{3} }$

$\: = \sf \: \dfrac{\dfrac{ - 5 + 3}{4} }{\dfrac{ - 5 - 4}{3} }$

$= \: \sf \: - \dfrac{ - 2}{4} \times \dfrac{3}{( - 9)}$

$\: = \sf \: \dfrac{1}{6}$

Additional Information :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ