If $\cot (x)=\dfrac{-5}{12}$, x lies in the second quadrant, find the values of the other five trigonometric functions.

Hint: In this question, we are given the value of the cotangent of the angle and the quadrant in which the angle x lies. Therefore, using the definition of cot(x), and other trigonometric formulas, we can obtain the values of other trigonometric ratios by solving the corresponding equations.

Complete step-by-step answer:
We are given the value of cot(x). We can use the trigonometric relation

$\text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\cot }^{2}}\left( x \right)$

With the given value of cot(x) to find

$\begin{align}

  & \text{cose}{{\text{c}}^{2}}\left( x \right)=1+{{\left( \dfrac{-5}{12}

\right)}^{2}}=1+\dfrac{25}{144}=\dfrac{144+25}{144}=\dfrac{169}{144} \\

 & \Rightarrow \text{cosec}(x)=\pm \sqrt{\dfrac{169}{144}}=\pm \dfrac{13}{12} \\

\end{align}$

However, in the second quadrant, the value of cosec(x) is positive, so we should take only the

positive value in the above equation to obtain

 $\text{cosec}(x)=\dfrac{13}{12}............(1.1)$

Also tan(x) and cot(x) are related by

$\tan (x)=\dfrac{1}{\cot (x)}$

Therefore, using the value of cot(x) in the above equation, we find

$\tan (x)=\dfrac{1}{\dfrac{-5}{12}}=\dfrac{-12}{5}................(1.2)$

We know that cosec(x) and sin(x) are related as

$\text{cosec}(x)=\dfrac{1}{\sin (x)}\Rightarrow \sin (x)=\dfrac{1}{\text{cosec}(x)}.........(1.3)$

Therefore, using the value of (1.1) in (1.3), we obtain

$\sin (x)=\dfrac{1}{\text{cosec}(x)}=\dfrac{1}{\dfrac{13}{12}}=\dfrac{12}{13}..........(1.4)$

Also, we know that

$\tan (x)=\dfrac{\sin (x)}{\cos (x)}\Rightarrow \cos (x)=\dfrac{\sin (x)}{\tan (x)}$

Therefore, using equations (1.2) and (1.4), we obtain

$\cos (x)=\dfrac{\dfrac{12}{13}}{\dfrac{-12}{5}}=\dfrac{-5}{13}..................(1.5)$

Also, sec(x) is given by

$\sec (x)=\dfrac{1}{\cos (x)}=\dfrac{1}{\dfrac{-5}{13}}=\dfrac{-13}{5}.............(1.6)$

Therefore, from equations (1.1), (1.2), (1.4), (1.5) and (1.5), we have found the other trigonometric ratios as

$\sin (x)=\dfrac{12}{13}$, $\cos (x)=\dfrac{-5}{13}$, $\tan (x)=\dfrac{-12}{5}$, $\sec (x)=\dfrac{-13}{5}$, $\text{cosec}(x)=\dfrac{13}{12}$

Which is the required answer to this question.



Note: We could also have found tan(x) by using $\tan (x)=\dfrac{1}{\cot (x)}$ and then sec(x) from tan(x) by using the identity ${{\sec }^{2}}\left( x \right)=1+{{\tan }^{2}}\left( x \right)$ and from there we could have found sin(x) and cos(x). However, the answer would have remained the same as found out in the solution above.