Question

What is $\cot A+\csc A$ equal to?
A. $\tan \dfrac{A}{2}$
B. $\cot \dfrac{A}{2}$
C. $2\tan \dfrac{A}{2}$
D. $2\cot \dfrac{A}{2}$

Answer 1

Use the facts that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ , $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ .
Use the following identities to convert from sum to product form:
$\sin 2A=2\sin A\cos A$
$\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)$
Recall that $\cos 0=1$ .

Complete step-by-step answer:
Using $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ , we can write:
$\cot A+\csc A$
= $\dfrac{\cos A}{\sin A}+\dfrac{1}{\sin A}$
Since $\cos 0=1$ , we can write it as:
= $\dfrac{\cos A+\cos 0}{\sin A}$
Using $\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)$ and $\sin 2A=2\sin A\cos A$ , we get:
= $\dfrac{2\cos \left( \tfrac{A}{2}+\tfrac{0}{2} \right)\cos \left( \tfrac{A}{2} - \tfrac{0}{2} \right)}{2\sin \tfrac{A}{2}\cos \tfrac{A}{2}}$
= $\dfrac{2\cos \tfrac{A}{2}\cos \tfrac{A}{2}}{2\sin \tfrac{A}{2}\cos \tfrac{A}{2}}$
Cancelling the common factors in the numerator and the denominator, we get:
= $\dfrac{\cos \tfrac{A}{2}}{\sin \tfrac{A}{2}}$
Which can also be written as:
= $\cot \dfrac{A}{2}$

The correct answer is, therefore, A. $\cot \dfrac{A}{2}$ .

Note: Angle Sum formula:
$\sin (A\pm B)=\sin A\cos B\pm \sin B\cos A$
$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$
Sum-Product formula:
$\sin 2A+\sin 2B=2\sin (A+B)\cos (A-B)$
$\sin 2A-\sin 2B=2\cos (A+B)\sin (A-B)$
$\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)$
$\cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B)$