Question

Question 5
Given $sec\theta$ = $\frac{13}{12}$, calculate all other trigonometric ratios.

Answer 1

Let $\Delta$POM be a right-angled triangle, right-angled at M.
We know that sec $\theta$ = $\frac{OP}{OM}$ = $\frac{13}{12}$
Let OP be 13k and OM will be 12k where k is a positive real number.

By Pythagoras theorem; we get,
$O{P}^2=O{M}^2+M{P}^2$
$(13k{)}^2=(12k{)}^2+M{P}^2$
$169k^2-144k^2=M{P}^2$
$M{P}^2=25k^2$
MP = 5k
Now,
$sin\theta$ = $\frac{MP}{OP}$ = $\frac{5k}{13k}$ = $\frac{5}{13}$
$cos\theta$ = $\frac{OM}{OP}$ = $\frac{12k}{13k}$ = $\frac{12}{13}$
$tan\theta$ = $\frac{MP}{OM}$ = $\frac{5k}{12k}$ = $\frac{5}{12}$
$cot\theta$ = $\frac{OM}{MP}$ = $\frac{12k}{5k}$ = $\frac{12}{5}$
$cosec\theta$ = $\frac{OP}{MP}$ = $\frac{13k}{5k}$ = $\frac{13}{5}$