താഴെ പറയുന്നവയിൽ ഏതാണ് സമവാക്യത്തിന്റെ വലതുഭാഗം

\(\frac{\sin ^2 A}{1-\cos A}=\)

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RRB Group D 1 Sept 2022 Shift 1 Official Paper
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  1. cosec A
  2. tan A
  3. cos A + 1
  4. sec A + tan A

Answer (Detailed Solution Below)

Option 3 : cos A + 1
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Detailed Solution

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നൽകിയത്:

, \(\frac{\sin ^2 A}{1-\cos A}=x\) എന്ന് കരുതുക

ഉപയോഗിച്ച സൂത്രവാക്യം:

  • sin2A = 1- cos2A
  • a2-b2=(a+b)(a-b)

കണക്കുകൂട്ടലുകൾ :

\(\frac{\sin ^2 A}{1-\cos A}=x\)

\( x = \frac{\sin ^2 A}{1-\cos A} = \frac{(1+cos A )(1-\cos A) }{(1-\cos A)} =1+cos A \)

x =cos A + 1

അതിനാൽ, \(\bf \frac{\sin ^2 A}{1-\cos A} =1+cos A \: \).

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