Question

Find the values of other five trigonometric functions
$\sec (x)=\dfrac{13}{5}$, x lies in the fourth quadrant,

Answer 1

Hint: In this question, we are given the value of the cosine of the angle and the quadrant in which the angle x lies. Therefore, using the definition of sec(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 know that cos(x) is the multiplicative inverse of sec(x). Therefore, using the given value of sec(x), we find

$\cos (x)=\dfrac{1}{\sec (x)}=\dfrac{1}{\dfrac{13}{5}}=\dfrac{5}{13}...........(1.1)$

Thus, we now know the value of cos(x). We can use the trigonometric relation

${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$

With the value of cos(x) from (1.1), we obtain

$\begin{align}

  & {{\sin }^{2}}\left( x \right)=1-{{\left( \dfrac{5}{13} \right)}^{2}}=1-\dfrac{25}{169}=\dfrac{169-25}{169}=\dfrac{144}{169} \\

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

\end{align}$

However, in the fourth quadrant, the value of sin(x) is negative, so we should take only the negative value in the above equation to obtain

 $\sin (x)=\dfrac{-12}{13}............(1.2)$

Now, the expression for tan(x) is given by,

\[\tan (x)=\dfrac{\sin (x)}{\cos (x)}\]

Therefore, using the values of cos(x) and sin(x) (1.1) and (1.2), we get

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

(x)}=\dfrac{\dfrac{-12}{13}}{\dfrac{5}{13}}=\dfrac{-12}{5}............(1.3)$

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

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

Therefore, using the value of tan(x) from (1.3) in the above equation, we find

$\cot (x)=\dfrac{1}{\dfrac{-12}{5}}=\dfrac{-5}{12}...............(1.4)$

We know that cosec(x) is the multiplicative inverse of sin(x), therefore using the value of

sin(x) from equation (1.2), we obtain

$\text{cosec}(x)=\dfrac{1}{\sin (x)}=\dfrac{1}{\dfrac{-12}{13}}=\dfrac{-13}{12}..........(1.5)$


Therefore, from equations (1.1), (1.2), (1.3), (1.4) and (1.5), we have found the other trigonometric ratios as

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

Which is the required answer to this question.



Note: We could also have found tan(x) from sec(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. Also, we should note that as x lies in the fourth quadrant, the value of sin(x) is negative but that of cos(x) is positive.