• # question_answer         If tan $\theta =\frac{b}{a}$then the value of $a.cos2\theta +b.sin2\theta$ is A) a                     B) b              C) $a-b$                           D) $a+b$

[a] $a.cos2\theta +b.sin2\theta$ $=a.\frac{1-tan\theta }{1+ta{{n}^{2}}\theta }+b\frac{2tan2\theta }{1+ta{{n}^{2}}\theta }=\frac{a(1-ta{{n}^{2}}\theta )+2btan\theta }{1+ta{{n}^{2}}\theta }$ $=\frac{a\left( 1-\frac{{{b}^{2}}}{{{a}^{2}}} \right)+2b.\frac{b}{a}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{a({{a}^{2}}-{{b}^{2}})+2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$ $=\frac{{{a}^{3}}-a{{b}^{2}}+2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$ $=\frac{{{a}^{3}}+a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=a$ Hence, option [a] is correct.