Question

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

Answer 1

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

$cot\,x\,=\frac{1}{sec\,x}=\frac{1}{(\frac{13}{5})}=\frac{5}{13}$

$⇒sin^2x+cos^2x=1$

$⇒sin^2x=1-cos^2x$

$⇒sin^2x=1-(\frac{5}{13})^2$

$⇒sin^2x=1-\frac{25}{169}=\frac{144}{169}$

$⇒sin\,x=±\frac{12}{13}$

Since x lies in the 4th quadrant, the value of sin x will be negative.

$sin\,x=-\frac{12}{13}$

$cosec\,x=\frac{1}{sin\,\,x}=\frac{1}{(-\frac{12}{13})}=-\frac{13}{12}$

$tax\,\,x=\frac{sin\,x}{cos\,\,x}=\frac{(\frac{-12}{13})}{(\frac{5}{13})}=-\frac{12}{5}$

$cot\,x=\frac{1}{tan\ x}=\frac{1}{(-\frac{12}{5})}=-\frac{5}{12}$