\(\dfrac{\left(cosa-sina\right)^2-\left(cosa+sina\right)^2}{cosa\cdot sina}\)

\(=\dfrac{\left(cosa-sina-cosa-sina\right)\left(cosa-sina+cosa+sina\right)}{cosa\cdot sina}\)

\(=\dfrac{-2\cdot sina\cdot2\cdot cosa}{cosa\cdot sina}=-4\)

\(\cos a-\sin a=\dfrac{1}{5}\\ \Leftrightarrow\left(\cos a-\sin a\right)^2=\dfrac{1}{25}\\ \Leftrightarrow1-2\sin a\cos a=\dfrac{1}{25}\\ \Leftrightarrow2\sin a\cos a=\dfrac{24}{25}\)

Mà \(\cos a=\dfrac{1}{5}+\sin a\)

\(\Leftrightarrow2\sin a\left(\dfrac{1}{5}+\sin a\right)=\dfrac{24}{25}\\ \Leftrightarrow\dfrac{2}{5}\sin a+2\sin^2a-\dfrac{24}{25}=0\\ \Leftrightarrow\left[{}\begin{matrix}\sin a=\dfrac{3}{5}\\\sin a=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\cos a=\dfrac{4}{5}\\\cos a=-\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\cot a=\dfrac{4}{5}\cdot\dfrac{5}{3}=\dfrac{4}{3}\)

Ta có: \(sin^2\alpha+cos^2\alpha=1\Rightarrow sin^2\alpha+\left(sin\alpha+\dfrac{1}{5}\right)^2=1\)

\(\Rightarrow25sin^2\alpha+5sin\alpha-12=0\\\Rightarrow\left(5sin\alpha-3\right)\left(5sin\alpha+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}sin\alpha=\dfrac{3}{5}\Rightarrow cos\alpha=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\Rightarrow cot\alpha=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\sin\alpha=-\dfrac{4}{5}\left(loại\right)\end{matrix}\right. \)

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=1\)

\(\Rightarrow a+b=45^0\)