Bài 2: 

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

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

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

a/ \(\sin\alpha=\frac{C_đ}{C_h}\)

\(\cos\alpha=\frac{C_k}{C_h}\)

\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{C_đ}{C_h}}{\frac{C_k}{C_h}}=\frac{C_đ}{C_k}=\tan\alpha\)

b/ \(\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{C_k}{C_h}}{\frac{C_đ}{C_h}}=\frac{C_k}{C_đ}=\cot\alpha\)

c/ \(\tan\alpha.\cot\alpha=\frac{C_đ}{C_k}.\frac{C_k}{C_đ}=1\)

d/ \(\sin^2\alpha=\frac{C_đ^2}{C_h^2}\)

\(\cos^2\alpha=\frac{C_k^2}{C_h^2}\)

\(\Rightarrow\sin^2\alpha+\cos^2\alpha=\frac{C_đ^2+C_k^2}{C_h^2}=\frac{C_h^2}{C_h^2}=1\)

P/s: hok trc lp 9 hay sao mà lm bài bài này?

\(C=\left(1+\tan^2\alpha\right).\cos^2\alpha+\left(1+\cot^2\alpha\right).\sin^2\alpha\)

\(=\cos^2\alpha+\cos^2\alpha.\tan^2\alpha+\sin^2\alpha+\sin^2\alpha.\cot^2\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\cos^2\alpha\right)\)

\(=1+1=2\)

Em dùng công thức sau 1+tan²x=\(\frac{1}{cos^2}\);\(1+cotg^2x=\frac{1}{sin^2x}\)với sin²x+cos²x=1

Đặt \(\left(sin^2\alpha;cos^2\alpha\right)=\left(a;b\right)\)=>1+a²=\(\frac{1}{b^2}\);\(1+b=\frac{1}{a^2}\);a²+b²=1

Suy ra C=\(\frac{1}{b^2}.\left(1-a^2\right)\)+\(\frac{1}{a^2}.\left(1-b^2\right)\)=\(\frac{1}{b^2}.b^2\)+\(\frac{1}{a^2}.a^2\)=2

Vậy C=2

\(\left(cosa-sina\right)^2=\frac{1}{25}\Leftrightarrow sin^2a+cos^2a-2sina.cosa=\frac{1}{25}\)

\(\Leftrightarrow\frac{sin^2a+cos^2a-2sina.cosa}{sin^2a}=\frac{1}{5sin^2a}=\frac{sin^2a+cos^2a}{5sin^2a}\)

\(\Leftrightarrow1+cot^2a-2cota=\frac{1}{5}+\frac{1}{5}cot^2a\)

\(\Leftrightarrow4cot^2a-10cota+4=0\Rightarrow\left[{}\begin{matrix}cota=2\\cota=\frac{1}{2}\end{matrix}\right.\)

Thui vậy! OK anh e sẽ giúp! Mà hok trc lp 9 hay sao mà chăm dữ?!

Có \(\cos\alpha-\sin\alpha=\frac{1}{5}\Rightarrow\left(\cos\alpha-\sin\alpha\right)^2=\frac{1}{25}\)

\(\Leftrightarrow\cos^2\alpha-2\sin\alpha.\cos\alpha+\sin^2\alpha=\frac{1}{25}\)

\(\Leftrightarrow1-2\sin\alpha.\cos\alpha=\frac{1}{25}\)

\(\Leftrightarrow\sin\alpha.\cos\alpha=\frac{12}{25}\Leftrightarrow\sin\alpha=\frac{12}{25\cos\alpha}\)

Thay vào biểu thức ban đầu rùi giải pt b2 là OK

@Ribi Nkok Ngok Bonking

Câu 1: 

\(\cos a=\sqrt{1-0.28^2}=\dfrac{24}{25}\)

\(\tan a=\dfrac{0.28}{0.96}=\dfrac{7}{24}\)

\(\cot a=\dfrac{1}{\tan a}=\dfrac{24}{7}\)

Câu 1: 

\(1+\cot^2a=\dfrac{1}{\sin^2a}\)

nên \(\dfrac{1}{\sin^2a}=1+5^2=26\)

\(\Leftrightarrow\sin^2a=\dfrac{1}{26}\)

\(\Leftrightarrow\sin a=\dfrac{\sqrt{26}}{26}\)

\(\cos a=\sqrt{1-\dfrac{1}{26}}=\dfrac{5\sqrt{26}}{26}\)

\(A=\dfrac{\sin a+\cos a}{\sin a-\cos a}=\left(\dfrac{\sqrt{26}+5\sqrt{26}}{26}\right):\left(\dfrac{\sqrt{26}-5\sqrt{26}}{26}\right)\)

\(=\dfrac{6\sqrt{26}}{-4\sqrt{26}}=\dfrac{-3}{2}\)