Find the value of cos 2A cos 2B + sin2(A - B) - sin2(A + B)

This question was previously asked in
SSC CGL 2022 Tier-I Official Paper (Held On : 12 Dec 2022 Shift 1)
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  1. sin (2A − 2B)
  2. sin (2A + 2B)
  3. cos (2A + 2B)
  4. cos (2A − 2B)

Answer (Detailed Solution Below)

Option 3 : cos (2A + 2B)
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SSC CGL Tier 1 2025 Full Test - 01
100 Qs. 200 Marks 60 Mins

Detailed Solution

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Given:

cos 2A cos 2B + sin2(A - B) - sin2(A + B)

Concept used:

cos (a + b) = cos a cos b - sin a sin b

sin2a - sin2b = sin(a + b) sin(a - b)

Calculation:

cos 2A cos 2B + sin2(A - B) - sin2(A + B)

⇒ cos 2A cos 2B - [sin2(A + B) - sin2(A - B)] 

{sin2a - sin2b = sin(a + b) sin(a - b)}

⇒ cos 2A cos 2B - [sin(A + B + A - B) sin(A + B - A + B)]

⇒ cos 2A cos 2B - [sin(A + A) sin(B + B)]

⇒ cos 2A cos 2B - sin 2A sin 2B

⇒ cos (2A + 2B)

∴ The required answer is cos (2A + 2B).

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