Question

Find the value of other five trigonometric functions if sec x = $\frac{13}{5}$, and x lies in fourth quadrant.

Answer 1

Here sec x = $\frac{13}{5}$

Now $se{c}^{2}x=1+ta{n}^2$ x

$\Rightarrow {\left(\frac{13}{5}\right)}^2=1+ta{n}^{2}x$

$\Rightarrow ta{n}^{2}x=\frac{169}{25}-1$

$\Rightarrow ta{n}^{2}x=\frac{144}{25}$

$\Rightarrow tanx=\pm \frac{12}{5}$

But x lies in fourth quadrant.

$\therefore tanx=\frac{-12}{5}$

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

cos x = $\frac{1}{\text{sec x}}=\frac{5}{13}$

Also $si{n}^{2}x+co{s}^{2}x=1$

$\Rightarrow si{n}^{2}x+{\left(\frac{5}{13}\right)}^2=1$

$\Rightarrow si{n}^{2}x=1-\frac{25}{169}$

$\Rightarrow si{n}^{2}x=\frac{144}{169}$

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

But x lies in fourth quadrant.

$\therefore sinx=\frac{-12}{13}$