Question
Without using trigonometric identity, show that:
$
\sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \cos e c 70^{\circ}=0
$
$
\sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \cos e c 70^{\circ}=0
$
✓
Answer
$
\begin{aligned}
& \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=0 \\
& \text { Consider } \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ} \\
& \Rightarrow \sec \left(90^{\circ}-20^{\circ}\right) \sin 20^{\circ}-\cos 20^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right) \\
& \Rightarrow \operatorname{cosec} 20^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \sec 20^{\circ} \\
& \Rightarrow \frac{1}{\sin 20^{\circ}} \cdot \sin 20^{\circ}-\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}} \\
& \Rightarrow 1-1=0
\end{aligned}
$
\begin{aligned}
& \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=0 \\
& \text { Consider } \sec 70^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \operatorname{cosec} 70^{\circ} \\
& \Rightarrow \sec \left(90^{\circ}-20^{\circ}\right) \sin 20^{\circ}-\cos 20^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right) \\
& \Rightarrow \operatorname{cosec} 20^{\circ} \sin 20^{\circ}-\cos 20^{\circ} \sec 20^{\circ} \\
& \Rightarrow \frac{1}{\sin 20^{\circ}} \cdot \sin 20^{\circ}-\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}} \\
& \Rightarrow 1-1=0
\end{aligned}
$
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