AP EAMCET · Maths · Trigonometric Equations
Let \(a\) and \(b\) be non-negative real numbers. If \(\sin x+a \cos\) \(\mathrm{x}=\mathrm{b}\), then \(|\mathrm{a} \sin \mathrm{x}-\cos \mathrm{x}|=\)
- A \(\sqrt{a^2-b^2+1}\)
- B \(\sqrt{b^2-a^2+1}\)
- C \(\sqrt{1+a^2+b^2}\)
- D \(\sqrt{a^2+b^2-1}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{a^2-b^2+1}\)
Step-by-step Solution
Detailed explanation
Given \(\sin \mathrm{x}+\mathrm{a} \cos \mathrm{x}=\mathrm{b}...(i)\) Let \(a \sin \mathrm{x}-\cos \mathrm{x}= \pm \alpha...(ii)\) on squanity and adding (i) and (ii) we got \(\mathrm{a}^2+1=\mathrm{b}^2+\alpha^2\) \[ \Rightarrow \alpha \sqrt{\mathrm{a}^2-\mathrm{b}^2+1} \]
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