giúp mình câu b) c) với ạ
b: \(sin^2b+cos^2b=1\)
=>\(sin^2b=1-\dfrac{1}{5}=\dfrac{4}{5}\)
=>\(sinb=\dfrac{2}{\sqrt{5}}\) hoặc \(sinb=-\dfrac{2}{\sqrt{5}}\)
TH1: \(sinb=\dfrac{2}{\sqrt{5}}\)
\(tanb=\dfrac{2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=2\)
cot b=1/tanb=1/2
TH2: \(sinb=-\dfrac{2}{\sqrt{5}}\)
\(tanb=\dfrac{-2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=-2\)
cot b=1/tan b=-1/2
c: \(1+cot^2y=\dfrac{1}{sin^2y}\)
=>\(\dfrac{1}{sin^2y}=1+2=3\)
=>\(sin^2y=\dfrac{1}{3}\)
=>\(siny=\dfrac{1}{\sqrt{3}}\) hoặc \(siny=-\dfrac{1}{\sqrt{3}}\)
TH1: \(siny=\dfrac{1}{\sqrt{3}}\)
\(coty=\dfrac{cosy}{siny}\)
=>\(cosy=\dfrac{1}{\sqrt{3}}\cdot\left(-\sqrt{2}\right)=\dfrac{-\sqrt{2}}{\sqrt{3}}\)
\(tany=\dfrac{1}{coty}=\dfrac{-1}{\sqrt{2}}\)
TH2: \(siny=-\dfrac{1}{\sqrt{3}}\)
\(cosy=coty\cdot siny=\left(-\sqrt{2}\right)\cdot\dfrac{-1}{\sqrt{3}}=\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}\)
$tany=\frac{1}{coty}=\frac{-1}{\sqrt{2}}$
