Question:

If \( a \cos \theta + b \sin \theta = 4 \) and \( a \sin \theta - b \cos \theta = 3 \), then the value of \( a^2 + b^2 \) is:

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Using the identity: \[ \cos^2 \theta + \sin^2 \theta = 1 \] simplifies such expressions.
Updated On: Oct 27, 2025
  • \( 7 \)
  • \( 16 \)
  • \( 25 \)
  • \( 36 \)
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The Correct Option is C

Solution and Explanation

Squaring both equations and adding:
\[ (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = 4^2 + 3^2 \] \[ a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta + a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = 16 + 9 \] \[ a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = 25 \] Since \( \cos^2 \theta + \sin^2 \theta = 1 \): \[ a^2 + b^2 = 25 \]
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