यदि \(\rm \tan \frac{A}{2}=x\) है, तो x ज्ञात कीजिए।

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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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दिया गया है:

tan (A/2) = x

प्रयुक्त सूत्र:

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

गणना:

tan (A/2) = x

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

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

∴ सही विकल्प 3 है।

Alternate Method

हम जानते हैं: tan(A/2) = sin(A/2) / cos(A/2). 
 
Sin(A/2) को पाइथागोरस सर्वसमिका, sin²λ + cos²λ = 1 से प्राप्त किया जा सकता है, जिसका अर्थ है कि sinλ = √(1 - cos²λ). तो, sin(A/2) = √(1 - cos²(A/2)).
 
इसे समीकरण में रखने पर, हमें यह प्राप्त होता है
 
tan(A/2) = √(1 - cos²(A/2)) / cos(A/2) .......(i)
 
अब, अर्ध कोण सूत्र से,
 
cos(A/2) = √((1 + cosA) / 2) .......(ii)
 
समीकरण (i) में (ii) का मान रखने पर:-
 
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).
 
अतः विकल्प 3 सही उत्तर है।
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