Question

Express the following in terms of trigonometric ratios of angles lying between $0$ and ${45}^{\circ }$ :
$\sec {76}^{\circ }+\text{cosec}{52}^{\circ }$

Answer 1

We have,

$\sec ({90}^{\circ }-\theta )=\csc \theta$ and $\csc ({90}^0-\theta )=\sec \theta$

$\sec {76}^0=\csc ({90}^0-{14}^0)=\csc {14}^0$

$\csc {52}^0=\sec ({90}^0-{38}^0)=\sec {38}^0$

$\therefore \sec {76}^0+\csc {52}^0=\csc {14}^0+\sec {38}^0$