If \(\rm \tan \frac{A}{2}=x\), then find x.

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SSC CGL 2023 Tier-I Official Paper (Held On: 18 Jul 2023 Shift 1)
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  1. \(\rm \frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}}\)
  2. \(\rm \frac{\sqrt{1-\sin A}}{\sqrt{1+\cos A}}\)
  3. \(\rm \frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}\)
  4. \(\rm \frac{\sqrt{\cos A-1}}{\sqrt{1+\cos A}}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}\)
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Detailed Solution

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Given:

tan (A/2) = x

Formula used:

tan (A/2) = ± \(\rm \frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}\)

Calculation:

tan (A/2) = x

⇒ \(\rm \frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}\) = x

⇒ x = \(\rm \frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}\)

∴ The correct option is 3.
Alternate Method 

We know: tan(A/2) = sin(A/2) / cos(A/2).
 
Sin(A/2) can be derived by the Pythagorean Identity, sin²λ + cos²λ = 1, which means sinλ = √(1 - cos²λ). So, sin(A/2) = √(1 - cos²(A/2)).
 
Putting this into the equation, we get
 
tan(A/2) = √(1 - cos²(A/2)) / cos(A/2)  .......(i)
 
Now, from half angle formula, 
 
cos(A/2) = √((1 + cosA) / 2)  .......(ii)
 
value of (ii) put in equation (i):-
 
tan(A/2) = √[1 - √{(1 + cosA)²/ 4}] / √((1 + cosA) / 2)
 

tan(A/2) = √[1 - {(1 + cosA)/ 2}] / √((1 + cosA) / 2)

tan(A/2) = √((2 - 1 - cosA)/ 2) / √((1 + cosA) / 2)

tan(A/2) = √((1 - cosA)/ 2) / √((1 + cosA) / 2)

tan(A/2) = ±√(1 - cosA) / √(1 + cosA).
 
So, option 3 is the correct answer.
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