Question

If cot x = -5/12 , x lies in second quadrant , find the values of other five trigonometric functions.

Answer 1

SOLUTION :-

Given,

cot x = $\sf-\dfrac{5}{12}$

$\longrightarrow\sf tan \ x = \dfrac{1}{cos \ x}$

$\longrightarrow\bf tan \ x = -\dfrac{12}{5}$

We know that,

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

$\longrightarrow\sf sec^2 \ x = 1+\left(-\dfrac{12}{5}\right)^2 \\\\\longrightarrow sec^2 \ x = 1+\dfrac{144}{25} \\\\\longrightarrow sec^2 \ x = \dfrac{169}{25} \\\\\longrightarrow sec \ x = \pm \dfrac{13}{5}$

Since x lies in second quadrant, sec x will be negative.

$\bf\therefore sec \ x = - \dfrac{13}{5}$

$\sf\longrightarrow cos \ x = \dfrac{1}{sec \ x } \\\\\longrightarrow\bf cos \ x = -\dfrac{5}{13}$

We know that,

$\boxed{\sf sin \ x = tan \ x \ cos \ x }$

$\longrightarrow\sf sin \ x = \sf \left( -\dfrac{12}{5}\right)\times \left( - \dfrac{5}{13}\right) \\\\\longrightarrow\bf sin \ x = \dfrac{13}{12}$

$\longrightarrow \sf cosec \ x =\dfrac{1}{sin \ x } \\\\\longrightarrow\bf cosec \ x =\dfrac{13}{12}$

Answer 2

Step-by-step explanation:

if we know the trigono merric ratios then it will be easier