A) \[a\]
B) \[b\]
C) \[-a\]
D) \[-b\]
Correct Answer: A
Solution :
Given that\[\tan \theta =\frac{b}{a}\]. Now, \[a\cos 2\theta +b\sin 2\theta =a\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+b\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)\] Putting\[\tan \theta =\frac{b}{a}\], we get \[=a\left( \frac{1-\frac{{{b}^{2}}}{{{a}^{2}}}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}} \right)+b\left( \frac{2\frac{b}{a}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}} \right)=a\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)+b\left( \frac{2ba}{{{a}^{2}}+{{b}^{2}}} \right)\] \[=\frac{1}{({{a}^{2}}+{{b}^{2}})}\{{{a}^{3}}-a{{b}^{2}}+2a{{b}^{2}}\}=\frac{a({{a}^{2}}+{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}=a\].
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