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

Hint: 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) $