Bài 3:
a)
$cos∝-sin∝=\dfrac{1}{5} ↔ (cos∝-sin∝)^2=\dfrac{1}{25}$
$↔cos^2∝-2sin∝cos∝+sin^2∝=\dfrac{1}{25}$
$↔1-2sin∝cos∝=\dfrac{1}{25}$ ( Vì $sin^2∝+cos^2∝=1$ )
$↔2sin∝cos∝=\dfrac{24}{25}$
$↔sin∝cos∝=\dfrac{12}{25}$ $(1)$
Ta có : $cos∝-sin∝=\dfrac{1}{5} → cos∝=sin∝+\dfrac{1}{5}$
$(1) ↔ sin∝.\left ( sin∝+\dfrac{1}{5} \right ) - \dfrac{12}{25}=0$
$↔sin^2∝+\dfrac{1}{5}sin∝-\dfrac{12}{25}=0$
$↔25sin^2∝+5sin∝-12=0$ ( Quy đồng khử mẫu )
$↔25sin^2∝+20sin∝-15sin∝-12=0$
$↔5sin∝(5sin∝+4)-3(5sin∝+4)=0$
$↔(5sin∝-3)(5sin∝+4)=0$
$↔\left [ \begin{array}{1}sin∝=\dfrac{3}{5}\\sin∝=-\dfrac{4}{5}\end{array} \right.↔\left [ \begin{array}{1}cos∝=\dfrac{4}{5}\\cos∝=-\dfrac{3}{5}\end{array} \right.$
$⇒cot∝=\dfrac{cos∝}{sin∝}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}$
Vậy $cot∝=\dfrac{4}{3}$
b)
Vì $ΔABC$ vuông tại $C$
Nên $sinB=cosA=\dfrac{5}{13}$
Ta có : $sin^2B+cos^2B=1$
$↔\dfrac{25}{169}+cos^2B=1$
$↔cos^2B=\dfrac{144}{169}$
$↔cosB=\dfrac{12}{13}$
$⇒tanB=\dfrac{sinB}{cosB}=\dfrac{5}{13}:\dfrac{12}{13}=\dfrac{5}{12}$
Vậy $tanB=\dfrac{5}{12}$
