\(cot 18^{\circ}\left(\cot 72^{\circ} \cdot \cos^2 22^{\circ} + \frac{1}{\tan 72^{\circ} \cdot \sec^2 68^{\circ}}\right)\) चे मूल्य काय आहे?

  1. 2
  2. 0
  3. 1
  4. यापैकी काहीही नाही

Answer (Detailed Solution Below)

Option 3 : 1
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Detailed Solution

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संकल्पना:

  • cot A = 1/ tan A      
  • cos A = 1/ sec A
  • sin A = cos (90° - A)
  • cos2 A + sin2 A = 1
  • tan A = cot (90° - A)

गणना:

\(cot 18^{\circ}\left(\cot 72^{\circ} \cdot \cos^2 22^{\circ} + \frac{1}{\tan 72^{\circ} \cdot \sec^2 68^{\circ}}\right)\)     \(cot 18^{\circ}\left(\cot 72^{\circ} \cdot \cos^2 22^{\circ} + \frac{1}{\tan 72^{\circ} \cdot \sec^2 68^{\circ}}\right)\)

⇒ cot 18° (cot 72° . cos2 22° + cot 72° . cos2 68°)      (∵ cot A = 1/ tan A and cos A = 1/ sec A)

⇒ cot 18° [cot 72° (cos2 22° + cos2 68°)]

⇒ cot 18° . cot 72° (cos2 22° + sin2 22°)      [∵ sin A = cos (90° - A)]

⇒ cot 18° . cot 72°      (∵ cos2 A + sin2 A = 1)

⇒ cot 18° . tan 18°      [∵ tan A = cot (90° - A)]

⇒ 1

म्हणून, \(cot 18^{\circ}\left(\cot 72^{\circ} \cdot \cos^2 22^{\circ} + \frac{1}{\tan 72^{\circ} \cdot \sec^2 68^{\circ}}\right)\)= 1.

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