Question

Cot x 3/4 x lies in 3rd quadrant

Answer 1

$\huge \underline {Question: -}$

• Cot x 3/4 x lies in 3rd quadrant

$\huge \underline {Given: -}$

• $\sf \cot(x) = \dfrac{3}{4}$ x lies in third quadrant

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

• Find all trigonometric functions

$\huge \underline {Solution : -}$

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

$\implies\sf\tan^{2} x = {( \dfrac{4}{3} )}^{2} = \dfrac{16}{9}$

$\implies\sf{ \sec }^{2} x - 1 = { \tan }^{2} x$

$\implies\sf{ \sec }^{2} x = { \tan }^{2} x + 1$

$\implies\sf= \dfrac{16}{9} + 1$

$\implies\sf= \dfrac{16 + 9}{9}$

$\implies\sf{ \sec }^{2} x = \dfrac{25}{9}$

$\implies\sf\sec(x) = \sqrt{ \dfrac{25}{9} } = \dfrac{ - 5}{3}$

$\implies\sf\cos(x) = \dfrac{1}{ \sec(x) } = \dfrac{ - 3}{5}$

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

$\implies\sf=1 - \dfrac{9}{25}$

$\implies\sf= \dfrac{25 - 9}{25}$

$\implies\sf{ \sin}^{2} x = \dfrac{16}{25}$

$\implies\sf\sin(x) = \sqrt{ \dfrac{16}{25} }$

$\implies\sf\sin(x) = - \dfrac{4}{5}$

$\implies\sf\csc(x) = \dfrac{1}{ \sin(x) } = \dfrac{ - 5}{4}$