Question

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

Answer 1

Hint: Start by finding the value of secx using the relation that the difference of squares of secant and tangent function is equal to 1. Now once you have got the value of secx, you can easily find other trigonometric ratios using the relation between the trigonometric ratios.

Complete step-by-step answer:
We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x.$ So, if we put the value of tanx in the formula, we get
$se{{c}^{2}}x=1+{{\left( -\dfrac{5}{12} \right)}^{2}}$
$\Rightarrow se{{c}^{2}}x=1+\dfrac{25}{144}$
$\Rightarrow {{\sec }^{2}}x=\dfrac{169}{144}$
Now we know that ${{a}^{2}}=b$ implies $a=\pm \sqrt{b}$ . So, our equation becomes:
$\Rightarrow \sec x=\pm \sqrt{\dfrac{169}{144}}=\pm \dfrac{13}{12}$
Now, as it is given that x lies in the second quadrant and secx is negative in the second quadrant.
$\therefore secx=-\dfrac{13}{12}$
Now we know that cosx is inverse of secx.
$\therefore \cos x=\dfrac{1}{\sec x}=-\dfrac{12}{13}$
Now using the property that tanx is the ratio of sinx to cosx, we get
$\tan x=\dfrac{\sin x}{\cos x}$
$\Rightarrow \tan x\cos x=\sin x$
$\Rightarrow \sin x=-\dfrac{5}{12}\times \left( -\dfrac{12}{13} \right)=\dfrac{5}{13}$
Now we know that cotx is the inverse of tanx. So, we can conclude that:
$\cot x=\dfrac{1}{\tan x}=-\dfrac{12}{5}$
Also, we know that secx is the inverse of cosx, and cosecx is the inverse of sinx.
$\therefore \cos ecx=\dfrac{1}{\sin x}=\dfrac{5}{13}$
$\therefore secx=\dfrac{1}{\cos x}=-\dfrac{13}{12}$

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, you need to remember the properties related to complementary angles and trigonometric ratios.