If acot θ = b, then what will be the value of \(\frac{bcosθ - asinθ}{bcosθ + asinθ}\)?

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SSC CGL 2023 Tier-I Official Paper (Held On: 24 Jul 2023 Shift 4)
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  1. \(\frac{b^2 + a^2}{b^2 - a^2}\)
  2. b2 + a2
  3. \(\frac{b^2 - a^2}{b^2 + a^2}\)
  4. 0

Answer (Detailed Solution Below)

Option 3 : \(\frac{b^2 - a^2}{b^2 + a^2}\)
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Detailed Solution

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Given

a cot θ = b

Calculation

\(\frac{bcosθ - asinθ}{bcosθ + asinθ}\)

Dividing the whole equation by cosθ, we get:

⇒ (b - asinθ/cosθ)/(b + asinθ/cosθ)

⇒ (b - atanθ)/(b + atanθ)

⇒ (b - a × a/b)/(b + a × a/b) 

⇒ (b2 - a2)/(b2 + a2) 

The value of the expression is  (b2 - a2)/(b2 + a2) .

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