a, Tìm được sinα = 24 5 , tanα = 24 , cotα =  1 24

b, cosα = 5 3 , tanα = 2 5 , cotα =  5 2

c, sinα = ± 2 5 , cosα = ± 1 5 , cotα =  1 2

d, sinα = ± 1 10 , cosα = ± 3 10 , tanα = 1 3

Bài 1: 

\(\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}\)

\(\tan\alpha=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\)

Bài 2: 

\(\sin\alpha=\sqrt{1-\dfrac{49}{100}}=\dfrac{\sqrt{51}}{10}\)

\(\tan\alpha=\dfrac{\sqrt{51}}{7}\)

⇔ cosα + sinα = 5(cosα - sinα)

⇔ cosα + sinα = 5cosα - 5sinα

⇔ 6sinα = 4cosα

a: \(\cos\alpha=\dfrac{1}{2}\)

\(\tan\alpha=\sqrt{3}\)

\(\cot\alpha=\dfrac{\sqrt{3}}{3}\)

\(A=\dfrac{\dfrac{sina}{cosa}+\dfrac{cosa}{cosa}}{\dfrac{sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{tana+1}{tana-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\)

3/4pi<a<pi

=>sin a>0; cosa<0

sin2a=-4/5

=>2*sina*cosa=-4/5

=>sina*cosa=-2/5

(sina-cosa)^2=sin^2a+cos^2a-2*sina*cosa=1+4/5=9/5

=>sin a-cosa=3/căn 5

Ta có  sin α + cos α = a ⇒ sin α + cos α 2 = a 2

⇔ sin 2 α + 2 sin α cos α + ​ c os 2 α = a 2 ⇔ 1 + 2 sin α cos α = a 2 ⇔ sin α cos α = a 2 − 1 2 .

Chọn C.

Ta có:

\(cot\alpha\cdot tan\alpha=1\)

\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}\)

\(\Rightarrow cota=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

Mà:

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

\(\Rightarrow sin\alpha=\sqrt{\dfrac{1}{cot^2\alpha+1}}\)

\(\Rightarrow sin\alpha=\sqrt{\dfrac{1}{\left(\dfrac{4}{3}\right)^2+1}}=\dfrac{3}{5}\) 

Lại có:

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

\(\Rightarrow cos\alpha=\sqrt{1-sin^2a}\)

\(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

\(tan\alpha=\dfrac{3}{4}\\ \Rightarrow cot\alpha=1:\dfrac{3}{4}=\dfrac{4}{3}\)

Có:

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\\ \Rightarrow sin\alpha=\sqrt{1:\left(1+\left(\dfrac{4}{3}\right)^2\right)}=\dfrac{3}{5}\)

\(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)