Question

Find the value of other five trigonometric ratios:
$\tan x = - \dfrac{5}{{12}}$ , x lies in the second quadrant.

Answer 1

Hint: Given that x lies in the second quadrant. Now if x lies in the 2nd quadrant, only \[\sin x\] and \[cosecx\] is positive.
Also, note the following important formulae:
$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
${\sin ^2}x + {\cos ^2}x = 1$
\[{\sec ^2}x - {\tan ^2}x = 1\]
\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
 Now, the value of \[tanx\] is given. Therefore find the value of the other five trigonometric ratios with the help of aforementioned formulae.

Complete step-by-step answer:
Given, $\tan x = - \dfrac{5}{{12}}$
Therefore $\cot x = \dfrac{1}{{\tan x}} = - \dfrac{{12}}{5}$
\[\because {\sec ^2}x - {\tan ^2}x = 1\]
\[ \Rightarrow {\sec ^2}x = 1 + {\tan ^2}x\]
Taking square root on both the sides we get,
\[ \Rightarrow \sec x = \pm \sqrt {1 + {{\tan }^2}x} \]
On substituting the value of \[tanx\] we get,
\[ \Rightarrow \sec x = \pm \sqrt {1 + {{\left( { - \dfrac{5}{{12}}} \right)}^2}} = \pm \sqrt {1 + \dfrac{{25}}{{144}}} \]
As, x lies in the second quadrant, so the value of \[secx\] is negative,
 \[ \Rightarrow \sec x = - \dfrac{{13}}{{12}}\]
Therefore $\cos x = \dfrac{1}{{\sec x}} = - \dfrac{{12}}{{13}}$
Now,
$\tan x = \dfrac{{\sin x}}{{\cos x}} = - \dfrac{5}{{12}}$
$ \Rightarrow \sin x = \cos x \times \left( { - \dfrac{5}{{12}}} \right)$
On substituting the value of \[cosx\] we get,
$ \Rightarrow \sin x = \left( { - \dfrac{{1{2}}}{{13}}} \right) \times \left( { - \dfrac{5}{{1{2}}}} \right) = \dfrac{5}{{13}}$
Therefore, $\operatorname{cosecx} = \dfrac{1}{{\sin x}} = \dfrac{{13}}{5}$
Hence when $\tan x = - \dfrac{5}{{12}}$ and x lies in second quadrant, the other five trigonometric ratios are :
\[co\operatorname{t} x = - \dfrac{{12}}{5}\] , $\sin x = \dfrac{5}{{13}}$ , $\cos x = - \dfrac{{12}}{{13}}$, \[\sec x = - \dfrac{{13}}{{12}}\] and $\operatorname{cosecx} = \dfrac{{13}}{5}$

Note:Note the following important formulae:
1.$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
2.${\sin ^2}x + {\cos ^2}x = 1$
3.\[{\sec ^2}x - {\tan ^2}x = 1\]
4.\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
5.$\sin ( - x) = - \sin x$
6.$\cos ( - x) = \cos x$
7.$\tan ( - x) = - \tan x$
8.$\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
9.$\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
10.$\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }$
Sign convention: