\(tan\alpha=2\)
=>\(sin\alpha=2\cdot cos\alpha\)
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+4=5\)
=>\(cos^2a=\dfrac{1}{5}\)
\(A=sin^2a+3\cdot sin^2a\cdot cos^2a-3\cdot cos^2a\)
\(=4\cdot cos^2a+3\cdot\left(2\cdot cosa\right)^2\cdot cos^2a-3\cdot cos^2a\)
\(=cos^2a+12\cdot cos^4a\)
\(=cos^2a\left(12\cdot cos^2a+1\right)\)
\(=\dfrac{1}{5}\left(\dfrac{12}{5}+1\right)=\dfrac{1}{5}\cdot\dfrac{17}{5}=\dfrac{17}{25}\)
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