Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)

\(=tan^2x+1+tan^2x=1+2tan^2x\)

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

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

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

a: \(G=\left(1-\sin^2a\right)\cdot\cot^2a+1-cot^2a\)

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

\(=\frac{cos^4a-cos^2a+\sin^2a}{\sin^2a}=\frac{cos^2a\left(cos^2a-1\right)+\sin^2a}{\sin^2a}\)

\(=\frac{cos^2a\cdot\left(-\sin^2a\right)+\sin^2a}{\sin^2a}=-cos^2a+1=\sin^2a\)

b: \(E=\frac{1-\sin^2a}{2\cdot\sin a\cdot cosa}=\frac{cos^2a}{2\cdot\sin a\cdot cosa}=\frac{cosa}{2\cdot\sin a}\)

c: \(P=cotx+\frac{\sin x}{1+cosx}\)

\(=\frac{cosx}{\sin x}+\frac{\sin x}{1+cosx}=\frac{cos^2x+cosx+sin^2x}{\sin x\left(1+cosx\right)}=\frac{cosx+1}{\sin x\left(1+cosx\right)}=\frac{1}{\sin x}\)

a: Xét ΔABC có \(0<\hat{A}<180^0;0<\hat{B}<180^0;0<\hat{C}<180^0\)

nên sin A>0; sin B>0; sin C>0

=>M=sinA+sinB+sinC>0

b: Xét ΔABC có \(\hat{A}>90^0\)

nên \(\hat{B}<90^0;\hat{C}<90^0\)

=>cos B>0; cosC>0; cosA<0

=>\(M=cosA\cdot cosB\cdot cosC<0\)

d: \(\hat{A}>90^0\)

=>cot A<0

\(0<\hat{B}<90^0;0<\hat{C}<90^0\)

=>tan B>0; cot C>0

=>\(D=\cot A\cdot\tan B\cdot\cot C\) <0

a: sin a=2/3

=>cos^2a=1-(2/3)^2=5/9

=>\(cosa=\dfrac{\sqrt{5}}{3}\)

\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)

\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

b: cos a=1/5

=>sin^2a=1-(1/5)^2=24/25

=>\(sina=\dfrac{2\sqrt{6}}{5}\)

\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

c: cot a=1/tana=1/2

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

=>1/cos^2a=1+4=5

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)

bài này không có giới hạn góc sao tìm được bạn .

Lớp 9 nên coi như các góc này đều nhọn

a.

\(cosa=\sqrt{1-sin^2a}=\dfrac{15}{17}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{8}{15}\)

\(cota=\dfrac{1}{tana}=\dfrac{15}{8}\)

b.

\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)

\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)

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

a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)

\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)

\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)