A) \[a\]
B) \[b\]
C) \[a+b\]
D) \[{{a}^{2}}+{{b}^{2}}\]
Correct Answer: A
Solution :
Given, \[\tan x=\frac{b}{a}\] \[\therefore \] \[a\,\,\cos \,\,2x+b\,\,\sin \,\,2x\] \[=a\left( \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)+b\left( \frac{2\tan x}{1+{{\tan }^{2}}x} \right)\] \[=a\left( \frac{1-{{b}^{2}}/{{a}^{2}}}{1+{{b}^{2}}/{{a}^{2}}} \right)+b\left( \frac{2b/a}{1+{{b}^{2}}/{{a}^{2}}} \right)\] \[=a\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)+\frac{2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{3}}-a{{b}^{2}}+2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{3}}+a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=a\frac{({{a}^{2}}+{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}=a\]
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