\(tan40^0 = \alpha\) అయితే ,  \(\frac{tan320^0 - tan 310^0}{1 + tan320^0.tan 310^0}\). యొక్క విలువ ఎంత?

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SSC CGL 2022 Tier-I Official Paper (Held On : 13 Dec 2022 Shift 3)
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  1. \(\frac{1 - \alpha^2}{\alpha}\)
  2. \(\frac{1 + \alpha^2}{2\alpha}\)
  3. \(\frac{1 - \alpha^2}{2\alpha}\)
  4. \(\frac{1 +\alpha^2}{\alpha}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{1 - \alpha^2}{2\alpha}\)
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Detailed Solution

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ఇచ్చిన దత్తాంశం:

tan 40° = α 

ఉపయోగించిన సూత్రం:

Tan (A - B) = (tan A - tan B)/1 + tan A × tan B

tan (90° - θ) = cot θ 

cot θ × tan θ = 1

సాధన:

⇒ \(\frac{tan320^0 - tan 310^0}{1 + tan320^0.tan 310^0}\) = tan (320° - 310°)

⇒ tan 10° 

ఇప్పుడు, మనం వ్రాయవచ్చు:

Tan 10° = tan (50° - 40°)

⇒ [(tan 50° - tan 40°)/1 + (tan 50° × tan 40°)]

⇒ [tan (90° - 40°) - tan 40°/1 + tan (90° - 40°) × tan 40°]

⇒ cot 40° - tan 40°/1 + cot 40° × tan 40°

⇒ (1/α - α)/1 + 1

⇒ (1/α - α)/2

⇒ (1 - α2)/2α 

∴ సరైన సమాధానం (1 - α2)/2α.

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