Question

$\sin {81}^{\circ }+\tan {81}^{\circ }$, when expressed in terms of angles between $0^{\circ }$ and ${45}^{\circ }$, becomes

• A. $\sin 9^{\circ }+\cos 9^{\circ }$
• B. $\cos 9^{\circ }+\tan 9^{\circ }$
• C. $\sin 9^{\circ }+\tan 9^{\circ }$
• D. $\cos 9^{\circ }+\cot 9^{\circ }$

Answer 1

Answer: (D)

$\cos 9^{\circ }+\cot 9^{\circ }$

We know that,

$\sin A=\cos (90˚-A)$

$\tan A=\cot (90˚-A)$

$\sec A=\text{cosec}(90˚-A)$

Using these, we get

$\sin {81}^{\circ }=\sin ({90}^{\circ }-9^{\circ })=\cos 9^{\circ }$
$\tan {81}^{\circ }=\tan (90˚-9)=\cot 9^{\circ }$
$\therefore \sin {80}^{\circ }+\tan {81}^{\circ }=\cos 9^{\circ }+\cot 9^{\circ }$