Question
Without using trigonometric identity, show that :
$
\sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2
$

Answer

$
\begin{aligned}
& \sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2 \\
& \text { consider } \sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ} \\
& \Rightarrow \sin 42^{\circ} \sec \left(90^{\circ}-42^{\circ}\right)+\cos 42^{\circ} \operatorname{cosec}\left(90^{\circ}-42^{\circ}\right) \\
& \Rightarrow \sin 42^{\circ} \cdot \operatorname{cosec} 42^{\circ}+\cos 42^{\circ} \sec 42^{\circ} \\
& \Rightarrow \sin 42^{\circ} \cdot \frac{1}{\sin 42^{\circ}}+\cos 42^{\circ} \frac{1}{\cos 42^{\circ}} \\
& \Rightarrow 1+1=2
\end{aligned}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free