A) a
B) b
C) \[a-b\]
D) \[a+b\]
Correct Answer: A
Solution :
[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.
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