Đặt \(x=\alpha\)

a: \(\dfrac{1}{\cos^2x}=1+\tan^2x=1+\dfrac{1}{9}=\dfrac{10}{9}\)

nên \(\cos x=\dfrac{3\sqrt{10}}{10}\)

=>\(\sin x=\dfrac{\sqrt{10}}{10}\)

b: \(\dfrac{1}{\sin^2x}=1+\cot^2x=1+\dfrac{9}{16}=\dfrac{25}{16}\)

\(\Leftrightarrow\sin x=\dfrac{4}{5}\)

hay \(\cos x=\dfrac{3}{5}\)

Ta có: sin 2 α + c o s 2 α = 1

Suy ra:  sin 2 α = 1 - c o s 2 α = 1 - 0 , 8 2  = 1 – 0,64 = 0,36

Vì sin α > 0 nên sin α = 0 , 36 = 0,6

Suy ra: tg  α  = sin⁡ α /cos⁡ α  = 0,6/0,8 = 3/4 = 0,75

cotg  α  = 1/tg α  = 1/0,75 = 1,3333

Bài 1: 

Ta có: \(A=\sin^6\alpha+3\cdot\sin^2\alpha\cdot\cos^2\alpha+\cos^6\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3-3\cdot\sin^2\alpha\cdot\cos\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)

\(=1^3\)

=1

lấy 1 ở đâu để trừ đi \(sin^2\alpha\) ạ????

a, Ta có tổng các góc bằng 180^o

=> \(\widehat{P}=55^o\)

- Áp dụng tỉ số lượng giác :

\(\cos35=\dfrac{MN}{4}\)

\(\Rightarrow MN\approx3,277cm\)

\(\sin35=\dfrac{MP}{4}\)

\(\Rightarrow MP\approx2,294cm\)

b, Ta có : \(A=\dfrac{2\cos^2a-\cos^2a-\sin^2a}{\sin a+\cos a}=\dfrac{\left(\sin a+\cos a\right)\left(\cos a-\sin a\right)}{\sin a+\cos a}\)

\(=\cos a-\sin a\)

c, \(sin30< sin35< cos40< sin60< cos25\)

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