Question

cosx=-1/2,x lies in third quadrant​

Answer 1

$\huge\mathtt{Question:-}$

• cosx=-1/2,x lies in third quadrant

$\huge\mathtt{Given:-}$

• cosx=-1/2

$\huge\mathtt{To \: find:-}$

• x lies in third quadrant

$\huge\mathtt{Solution:-}$

$\implies\sf \sin^{2} x + \cos ^{2} x = 1$

$\implies\sf \sin ^{2} x = 1 - \cos^{2} x$

$\implies\sf \sin^{2} x =1 - \dfrac{1}{4}$

$\: \: \: \: \: \: \: \: \: \: = \dfrac{4 - 1}{4}$

$\implies\sf \sin^{2} x = \dfrac{3}{4}$

$\implies\sf \sin(x) = \sqrt{ \dfrac{3}{4} }$

$\boxed { \sf \sin(x) = \dfrac{ - \sqrt{3} }{2} }$

$\implies\sf \cos ^{2} x = {\big(} \frac{ - 1}{2}\big )^2$

$\implies\sf \cos^{2} x = \dfrac{1}{4} \sf \: \: (x \: lies \: in \: 3^{rd} \: quadrant)$

$\implies\sf \tan(x) = \sf \dfrac{ \sin( x) }{ \cos(x) } = \frac{ - \dfrac{ \sqrt{3} }{2} }{ - \dfrac{ 1}{2} } = \dfrac{ \sqrt{3} }{2} \times \dfrac{2}{1} = \sqrt{3}$

$\implies\sf \csc(x) = \dfrac{1}{ \sin(x) } = \dfrac{-2}{ \sqrt{3} }$

$\implies\sf \sec(x) = \dfrac{1}{ \cos(x) } = \dfrac{ - 2}{1} = - 2$

$\implies\sf \cot(x) = \dfrac{1}{ \sqrt{3} }$

Hence solved !!