Question

$\sin {84}^{\circ }+\sec {84}^{\circ }$ expressed in terms of angles between $0^{\circ }$ and ${45}^{\circ }$ becomes

• A. $\cos 6^{\circ }+\text{cosec}{6}^{\circ }$
• B. $\sin 6^{\circ }+\cos 6^{\circ }$
• C. $\sin 6^{\circ }+\text{cosec}{6}^{\circ }$
• D. None of these

Answer 1

The correct option is A $\cos 6^{\circ }+\text{cosec}{6}^{\circ }$

We know that

$\sin A=\cos (90˚-A)$

$\tan A=\cot (90˚-A)$

$\sec A=\text{cosec}(90˚-A)$

Using these, we get

$\sin {84}^{\circ }=\cos (90˚-84˚)=\cos 6^{\circ }$

$\sec {84}^{\circ }=\text{cosec}(90˚-84˚)=\text{cosec}{6}^{\circ }$
$\therefore \sin {84}^{\circ }+\sec {84}^{\circ }=\cos 6^{\circ }+\text{cosec}{6}^{\circ }$.