Question

Show that:

$\sin {42}^{\circ }\sec {48}^{\circ }+\cos {42}^{\circ }cosec{48}^{\circ }=2$

Answer 1

Consider LHS

$=\sin {42}^{\circ }\sec {48}^{\circ }+\cos {42}^{\circ }cosec{48}^{\circ }$

$=\sin ({90}^{\circ }-{48}^{\circ })\frac{1}{\cos {48}^{\circ }}+\cos ({90}^{\circ }-{48}^{\circ })\frac{1}{\sin {48}^{\circ }}$

$=\cos {48}^{\circ }\times \frac{1}{\cos {48}^{\circ }}+\sin {48}^{\circ }\times \frac{1}{\sin {48}^{\circ }}$

On simplifying we get,

= 1 + 1

= 2

= RHS