Bài 2: 

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

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

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

ta có : \(tan\alpha+cot\alpha=3\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}=3\)

\(\Leftrightarrow\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}=3\Leftrightarrow\dfrac{1}{sin\alpha.cos\alpha}=3\)

\(\Leftrightarrow sin\alpha.cos\alpha=\dfrac{1}{3}\) vậy \(sin\alpha.cos\alpha=\dfrac{1}{3}\)

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

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

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

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

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

Ta có:

 \(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi  + \alpha } \right) =  \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2}  + \alpha } \right) - \cos \left( {12\pi  + \pi + \alpha } \right) =  \sin \left( {-8\pi  + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha  \right) = \cos \left( \alpha  \right) + \cos \left( \alpha  \right) = 2\cos \left( \alpha  \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)

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?

\(\cos \alpha  =  - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}}  =  - \frac{{12}}{{13}}\) (vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\))

\(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)

\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha  + \sin \frac{\pi }{4}sin\alpha  = \frac{{ - 17\sqrt 2 }}{{26}}\)

\(1+\tan^2a=1+\frac{\sin^2a}{\cos^2a}=\frac{\sin^2a+\cos^2a}{\cos^2a}=\frac{1}{\cos^2a}\)

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

cot a=1/5 nên cosa/sina=1/5

=>sina=5cosa

\(1+cot^2a=\dfrac{1}{sin^2a}=1+\dfrac{1}{25}=\dfrac{26}{25}\)

nên \(sina=\dfrac{5}{\sqrt{26}}\Leftrightarrow cosa=\dfrac{1}{\sqrt{26}}\)

\(cot^4a+sin^2a-cos^2a\)

\(=\dfrac{1}{5^4}+25cos^2a-cos^2a\)

\(=\dfrac{1}{5^4}+24\cdot\dfrac{1}{26}=\dfrac{7513}{8125}\)