Q.If \( a\cos\theta + b\sin\theta = c \), then what is the value of \( a\sin\theta - b\cos\theta \)? (a) \(\pm\sqrt{a^2-b^2+c^2}\) (b) \(\pm\sqrt{a^2+b^2-c^2}\) (c) \(\pm\sqrt{a^2-b^2-c^2}\) (d) \(\pm\sqrt{b^2+c^2-a^2}\)
Answer: B
If \( a\cos\theta + b\sin\theta = c \) Then, \( (a\cos\theta + b\sin\theta)^2 = c^2 \) So, \( a^2\cos^2\theta + b^2\sin^2\theta + 2ab\cos\theta\sin\theta = c^2 \) Or, \( a^2(1 - \sin^2\theta) + b^2(1 - \cos^2\theta) + 2ab\sin\theta\cos\theta = c^2 \) Which becomes: \( a^2 - a^2\sin^2\theta + b^2 - b^2\cos^2\theta + 2ab\sin\theta\cos\theta = c^2 \) Or, \( -a^2\sin^2\theta - b^2\cos^2\theta + 2ab\sin\theta\cos\theta = c^2 - a^2 - b^2 \) Now, \( a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta = a^2 + b^2 - c^2 \) So, \( (a\sin\theta - b\cos\theta)^2 = a^2 + b^2 - c^2 \) Therefore, \( a\sin\theta - b\cos\theta = \pm\sqrt{a^2 + b^2 - c^2} \)
Similar Questions