Question

$\text{cosec}{69}^{\circ }+\cot {69}^{\circ }$, when expressed in terms of angles between $0^{\circ }$ and ${45}^{\circ }$, becomes

• A. $\sec {21}^{\circ }+\tan {21}^{\circ }$
• B. $\sin {21}^{\circ }+\cot {21}^{\circ }$
• C. $\sin {21}^{\circ }+\cos {21}^{\circ }$
• D. $\sec {21}^{\circ }+\cot {21}^{\circ }$

Answer 1

The correct option is A $\sec {21}^{\circ }+\tan {21}^{\circ }$

We know that:

$\cos A=\sin (90˚-A)$

$\cot A=\tan (90˚-A)$

$\text{cosec}A=\sec (90˚-A)$

Using these, we get:

$\text{cosec}{69}^{\circ }=\text{cosec}(90˚-21˚)=\sec 21˚$
$\cot {69}^{\circ }=\cot (90˚-21˚)=\tan {21}^{\circ }$

$\therefore \text{cosec}{69}^{\circ }+\cot {69}^{\circ }=\sec {21}^{\circ }+\tan {21}^{\circ }$