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\(\frac{1-tana}{1+tana}=\frac{1-\frac{sina}{cosa}}{1+\frac{sina}{cosa}}=\frac{\frac{1}{cosa}\left(cosa-sina\right)}{\frac{1}{cosa}\left(cosa+sina\right)}=\frac{cosa-sina}{cosa+sina}\)
\(\sin\alpha=\frac{2}{3}\) nên a là góc nhọn trong tam giác vuông có cạnh đối là 2, cạnh huyền là 3 suy ra cạnh kề = \(\sqrt{5}\)
Vậy: \(\cos\alpha=\sqrt{\frac{5}{3}};\tan\alpha=\frac{2}{\sqrt{5}};\cot\alpha=\sqrt{\frac{5}{2}}\)
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
D = \(\left(sin^2a+cos^2a\right)+\left(cos\left(90-a\right)-sina\right)+1+\left(tan^2\left(90-a\right)-\frac{1}{sin^2a}\right)\)
\(=1+\left(sina-sina\right)+1+\left(cot^2a-1-cos^2a\right)=1+1-1=1\)
\(\sin a.\cos a=\frac{\sqrt{3}}{4}\)
=> \(\sin a=\frac{\sqrt{3}}{4\cos a}\)
=> \(\frac{3}{16\cos^2a}+\cos^2a=1\)
=> \(16\cos^4a-16\cos^2a-3=0\)
=> \(\left[\begin{array}{nghiempt}\cos^2a=\frac{2+\sqrt{7}}{4}\Rightarrow\cos a=\pm\frac{\sqrt{2+\sqrt{7}}}{2}\\\cos^2a=\frac{2-\sqrt{7}}{4}\end{array}\right.\)