Question

Evaluate the following :
$sec{50}^{\circ }sin{40}^{\circ }+cos{40}^{\circ }cosec{50}^{\circ }$

Answer 1

We have,
$sec{50}^{\circ }sin{40}^{\circ }+cos{40}^{\circ }cosec{50}^{\circ }$

$=sec({90}^{\circ }-{40}^{\circ })sin{40}^{\circ }+cos{40}^{\circ }cosec({90}^{\circ }-{40}^{\circ })$

$=cosec{40}^{\circ }sin{40}^{\circ }+cos{40}^{\circ }sec{40}^{\circ }=\frac{sin{40}^{\circ }}{sin{40}^{\circ }}+\frac{cos{40}^{\circ }}{cos{40}^{\circ }}=1+1=2$