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Question: The value \( \tan \dfrac{A}{2} \) is equal to: A. \( \csc A+\cot A \) B. \( \csc A-\cot A \) ...

The value tan⁡A2\tan \dfrac{A}{2} is equal to:
A. csc⁡A+cot⁡A\csc A+\cot A
B. csc⁡A−cot⁡A\csc A-\cot A
C. sec⁡A+tan⁡A\sec A+\tan A
D. sec⁡A−tan⁡A\sec A-\tan A

Explanation

Solution

Hint : Recall that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } , cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } , sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } .
Use the formulae: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta , cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta and sin⁡2θ+cos⁡2θ=1{{\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⁡A2\tan \dfrac{A}{2} can be written as:
tan⁡A2=sin⁡A2cos⁡A2\tan \dfrac{A}{2}=\dfrac{\sin \tfrac{A}{2}}{\cos \tfrac{A}{2}}
Multiplying both the numerator and the denominator by 2sin⁡A22\sin \dfrac{A}{2} , we get:
⇒ tan⁡A2=2sin⁡2A22sin⁡A2cos⁡A2\tan \dfrac{A}{2}=\dfrac{2{{\sin }^{2}}\tfrac{A}{2}}{2\sin \tfrac{A}{2}\cos \tfrac{A}{2}}
Using the formulae sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta , cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta , we get:
⇒ tan⁡A2=1−cos⁡Asin⁡A\tan \dfrac{A}{2}=\dfrac{1-\cos A}{\sin A}
Using the definitions cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } , we can write it as:
⇒ tan⁡A2=csc⁡A−cot⁡A\tan \dfrac{A}{2}=\csc A-\cot A
The correct answer option is B. csc⁡A−cot⁡A\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⁡θ=PH\sin \theta =\dfrac{P}{H} , cos⁡θ=BH\cos \theta =\dfrac{B}{H} , tan⁡θ=PB\tan \theta =\dfrac{P}{B}
tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } , cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }
csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } , sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } , tan⁡θ=1cot⁡θ\tan \theta =\dfrac{1}{\cot \theta }
Angle Sum formula:
sin⁡(A±B)=sin⁡Acos⁡B±sin⁡Bcos⁡A\sin (A\pm B)=\sin A\cos B\pm \sin B\cos A
cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B
Sum-Product formula:
sin⁡2A+sin⁡2B=2sin⁡(A+B)cos⁡(A−B)\sin 2A+\sin 2B=2\sin (A+B)\cos (A-B)
sin⁡2A−sin⁡2B=2cos⁡(A+B)sin⁡(A−B)\sin 2A-\sin 2B=2\cos (A+B)\sin (A-B)
cos⁡2A+cos⁡2B=2cos⁡(A+B)cos⁡(A−B)\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)
cos⁡2A−cos⁡2B=−2sin⁡(A+B)sin⁡(A−B)\cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B)