Question

tan x= -5/12,x lies in second quadrant ​

Answer 1

Question :-

• tan x= -5/12,x lies in second quadrant

Given :-

• tan x= -5/12

To find :-

• Find the value of all trigonometric functions

Solution :-

$\longrightarrow \sf \tan(x) = \dfrac{ - 5}{12}$

$\longrightarrow \sf \cot(x) = \dfrac{1}{ \tan(x) } = \dfrac{ - 5}{12}$

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

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

$\: \: \: \: \: \: \: \: \: \: = \dfrac{25}{144} + 1$

$\: \: \: \: \: \: \: \: \: \: = \dfrac{25 + 144}{144}$

$\longrightarrow \sf{ \sec }^{2}x = \dfrac{169}{144}$

$\longrightarrow \sf \sec(x) = \sqrt{ \dfrac{169}{144} }$

$\longrightarrow \sf \sec(x) = \dfrac{ - 13}{12}$

$\longrightarrow \sf \cos(x) = \dfrac{1}{ \sec(x) } = \dfrac{ - 12}{13}$

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

$\: \: \: \: \: \: \: \: \: \sf = 1 - \dfrac{144}{169}$

$\: \: \: \: \: \: \: \: \: \sf = \dfrac{169 - 144}{169}$

$\longrightarrow \sf { \sin }^{2}x = \dfrac{25}{169}$

$\longrightarrow \sf \sin(x) = \sqrt{ \dfrac{25}{169} }$

$\longrightarrow \sf\sin(x) = \dfrac{5}{13}$

$\longrightarrow \sf \csc(x) = \dfrac{1}{ \sin(x) } = \dfrac{13}{5}$