Question

Find the value of other five trigonometric functions if tan x = $\frac{-5}{12}$, and x lies in second quadrant.

Answer 1

Here tan x = - $\frac{5}{12}$

cot x = $\frac{1}{\text{tan x}}=\frac{-12}{5}$

Now ${\text{sec}}^{2}x=1+{\text{tan}}^2$ x

$\Rightarrow se{c}^{2}x=1+{\left(\frac{-5}{12}\right)}^2$

$\Rightarrow se{c}^{2}x=\frac{169}{144}$

$\Rightarrow secx=\pm \frac{13}{12}$

But x lies in second quadrant.

$\therefore$ sec x= $\frac{-13}{12}$
cos x = $\frac{1}{\text{sec x}}=-\frac{-12}{13}$

Also ${\text{sin}}^{2}x+{\text{cos}}^2$ x= 1

$\to {\text{sin}}^{2}x+{\left(\frac{-12}{13}\right)}^2$ = 1

$\Rightarrow {\text{sin}}^{2}x=\frac{25}{169}$

$\Rightarrow {\text{sin}}^2$ x = $\pm \frac{5}{13}$

But x lies in second quadrant.

$\therefore$ sin x =$\frac{5}{13}$

cosec x = $\frac{1}{\text{sin x}}=\frac{13}{5}$.