Question

$\tan {68}^{\circ }+\sec {68}^{\circ }$, when expressed in terms of angles between $0^{\circ }$ and ${45}^{\circ }$, becomes

• A. $\cot {22}^{\circ }+\text{cosec}{22}^{\circ }$
• B. $\sin {22}^{\circ }+\cos {22}^{\circ }$
• C. $\cos {22}^{\circ }+\tan {22}^{\circ }$
• D. None of these

Answer 1

The correct option is A $\cot {22}^{\circ }+\text{cosec}{22}^{\circ }$

Given trigonometric expression is $\tan {68}^o+\sec {68}^o$

We know that,

$\sin A=\cos (90˚-A)$

$\tan A=\cot (90˚-A)$

$\sec A=\text{cosec}(90˚-A)$

Using these, we get:

$\tan {68}^{\circ }=\tan (90˚-22˚)=\cot {22}^{\circ }$
$\sec {68}^{\circ }=\text{sec}(90˚-22˚)=\text{cosec}22˚$

$\therefore \tan {68}^{\circ }+\sec {68}^{\circ }=\cot {22}^{\circ }+\text{cosec}{22}^{\circ }$.