\(\frac{{{{\tan }^2}60^\circ + {{\cot }^2}30^\circ }}{{{{\tan }^2}20^\circ - {{\sec }^2}20^\circ }} - \frac{{\sin 90^\circ + \cos 0^\circ }}{{\cos e{c^2}20^\circ - {{\cot }^2}20^\circ }}\) এর মান নির্ণয় করুন।

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RRB Group D 13 Sept 2022 Shift 1 Official Paper
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  1. - 8
  2. - 4
  3. 4
  4. - 2

Answer (Detailed Solution Below)

Option 1 : - 8
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Detailed Solution

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অনুসৃত সূত্র:

tan 60º = √3

cot 30º = √3

sec2a - tan2a = 1

cosec2a - cot2a = 1

গণনা:

\(\frac{{{{\tan }^2}60^\circ + {{\cot }^2}30^\circ }}{{{{\tan }^2}20^\circ - {{ \sec }^2}20^\circ }} - \frac{{\sin 90^\circ + \cos 0^\circ }}{{\cos e{c^2}20^\circ - {{\ খাট }^2}20^\circ }}\)

⇒ (√32 + √32)/(-1) - (1 + 1)/1

⇒ - 6 - 2 = - 8

সুতরাং, সঠিক উত্তর হল -8

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