\(0< a< \dfrac{pi}{2}\)

=>\(sina>0\)

\(sin^2a+cos^2a=1\)

=>\(sin^2a=1-\dfrac{16}{25}=\dfrac{9}{25}\)

=>\(sina=\dfrac{3}{5}\)

\(sin2a=2\cdot sina\cdot cosa=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{24}{25}\)

=>Chọn B

\(sin^2\alpha+cos^2\alpha=1\Rightarrow cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\frac{7}{25}\right)^2}=\frac{24}{25}\)

\(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}\)

\(cot\alpha=\frac{1}{tan\alpha}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)

Ta có

\(\sin^2\alpha+\cos^2\alpha=1\\ \Leftrightarrow\left(\frac{7}{25}\right)^2+\cos^2\alpha=1\\ \Leftrightarrow\frac{49}{625}+\cos^2\alpha=1\\ \Leftrightarrow\cos^2\alpha=\frac{576}{625}\\ \Leftrightarrow\cos\alpha=\frac{24}{25}\)

Từ đó suy ra

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{7}{25}:\frac{24}{25}=\frac{7}{24}\\ \cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)

Lớp 9 không biết có học tới sin cos âm chưa nếu chưa thì lấy phần dương nha 

\(1+tan^2a=\frac{1}{cos^2a}\)    

\(1+\left(\frac{2}{3}\right)^2=\frac{1}{cos^2a}\)    

\(1+\frac{4}{9}=\frac{1}{cos^2a}\)    

\(\frac{13}{9}=\frac{1}{cos^2a}\)    

\(cos^2a=\frac{9}{13}\)   

\(cosa=\pm\sqrt{\frac{9}{13}}=\pm\frac{3\sqrt{13}}{13}\)    

\(sin^2a+cos^2a=1\)   

\(sin^2a+\frac{9}{13}=1\)    

\(sin^2a=\frac{4}{13}\)    

\(sina=\pm\sqrt{\frac{4}{13}}=\pm\frac{2\sqrt{13}}{13}\)   

tan dương nên sẽ có 2 TH 

TH1 sin và cos cùng dương 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{2\sqrt{13}}{13}\right)^3+3\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}\)    

\(=-\frac{89}{459}\)   

TH2 sin và cos cùng âm 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{-2\sqrt{13}}{13}\right)^3+\left(\frac{-3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{-2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{-3\sqrt{13}}{13}\right)^3}\)

\(=-\frac{89}{459}\)

Bài 2: 

\(\cos a=\sqrt{1-\left(\dfrac{7}{25}\right)^2}=\dfrac{24}{25}\)

\(\tan a=\dfrac{7}{25}:\dfrac{24}{25}=\dfrac{7}{24}\)

\(\cot a=\dfrac{24}{7}\)

b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)

hay \(\cos\alpha=\dfrac{4}{5}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)

\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)

\(=\dfrac{141}{25}\)

c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)