Question

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

Answer 1

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

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

$1+tan^2x=sec^2x$

$⇒1+(-\frac{5}{12})^2=sec^2x$

$⇒1+\frac{25}{144}=sec^2x$

$⇒\frac{169}{144}=sec^2x$

$∴sec\,x=±\frac{13}{12}$

Since x lies in the 2nd quadrant, the value of sec x will be negative.

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

$cos\,x=\frac{1}{sec\,x}=\frac{1}{(-\frac{13}{12})}=-\frac{12}{13}$

$tan\,x=\frac{sin\,x}{cos\,\,x}$

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

$⇒sin\,x=(-\frac{5}{12})×(-\frac{12}{13})={\frac{5}{13}}$

$cosec\,x=\frac{1}{sin\,x}=\frac{1}{(\frac{5}{13})}=\frac{13}{5}$