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

To do:

We have to prove that $\sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=2$.

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

$cos\ (90^{\circ}- \theta) = sin\ \theta$

$cos\ \theta \times \sec\ \theta=1$

$sin\ \theta \times \operatorname{cosec}\ \theta=1$

Therefore,

$\sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=\sin (90^{\circ}- 42^{\circ})\sec 42^{\circ} + \cos (90^{\circ}- 42^{\circ})\operatorname{cosec} 42^{\circ}$

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

$=1+1$

$=2$

Hence proved.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

21 Views

Kickstart Your Career

Get certified by completing the course

Get Started