Question

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

Answer 1

Given $\sin {48}^{\circ }\sec {42}^{\circ }+\cos {48}^{\circ }\csc {42}^{\circ }$

we know that,

$\sec \theta =\frac{1}{\cos \theta }$ and $\csc \theta =\frac{1}{\sin \theta }$

$\therefore \frac{\sin {48}^{\circ }}{\cos {42}^{\circ }}+\frac{\cos {48}^{\circ }}{\sin {42}^{\circ }}$

We have,

$\cos (90-\theta )=\sin \theta ,\sin (90-\theta )=\cos \theta$

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

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

$=1+1$

$=2$

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

hence proved