Question

The value $ \tan \dfrac{A}{2} $ is equal to:
A. $ \csc A+\cot A $
B. $ \csc A-\cot A $
C. $ \sec A+\tan A $
D. $ \sec A-\tan A $

Answer 1

Hint: Recall that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ , $ \cot \theta =\dfrac{\cos \theta }{\sin \theta } $ , $ \sec \theta =\dfrac{1}{\cos \theta } $ and $ \csc \theta =\dfrac{1}{\sin \theta } $ .
Use the formulae: $ \sin 2\theta =2\sin \theta \cos \theta $ , $ \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta $ and $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ .
Get an expression in terms of only sin and cos, and simplify.

Complete step-by-step answer:
The trigonometric ratio $ \tan \dfrac{A}{2} $ can be written as:
  $ \tan \dfrac{A}{2}=\dfrac{\sin \tfrac{A}{2}}{\cos \tfrac{A}{2}} $
Multiplying both the numerator and the denominator by $ 2\sin \dfrac{A}{2} $ , we get:
⇒ $ \tan \dfrac{A}{2}=\dfrac{2{{\sin }^{2}}\tfrac{A}{2}}{2\sin \tfrac{A}{2}\cos \tfrac{A}{2}} $
Using the formulae $ \sin 2\theta =2\sin \theta \cos \theta $ , $ \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta $ , we get:
⇒ $ \tan \dfrac{A}{2}=\dfrac{1-\cos A}{\sin A} $
Using the definitions $ \cot \theta =\dfrac{\cos \theta }{\sin \theta } $ and $ \csc \theta =\dfrac{1}{\sin \theta } $ , we can write it as:
⇒ $ \tan \dfrac{A}{2}=\csc A-\cot A $
The correct answer option is B. $ \csc A-\cot A $
So, the correct answer is “Option B”.

Note: In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
  $ \sin \theta =\dfrac{P}{H} $ , $ \cos \theta =\dfrac{B}{H} $ , $ \tan \theta =\dfrac{P}{B} $
  $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ , $ \cot \theta =\dfrac{\cos \theta }{\sin \theta } $
  $ \csc \theta =\dfrac{1}{\sin \theta } $ , $ \sec \theta =\dfrac{1}{\cos \theta } $ , $ \tan \theta =\dfrac{1}{\cot \theta } $
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) $