Sửa đề: cosa=3/5

3pi/2<a<2pi

=>sin a<0

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

=>\(sin^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)

mà sin a<0

nên sina =-4/5

tan a=-4/5:3/5=-4/3

cot a=1:(-4/3)=-3/4

Ta có: \(\tan\alpha.\cot\alpha=1\Rightarrow\tan\alpha=\frac{1}{\cot\alpha}\)

Đặt \(\cot\alpha=t\)thì \(\tan\alpha=\frac{1}{t}\)

Khi đó \(B=\frac{1}{1+\frac{1}{t}}+\frac{1}{1+t}=\frac{t}{t+1}+\frac{1}{1+t}=1\)

1+tan a=1+sina/cosa = sina+cosa/cosa

1+cota=sina+cosa/sina

=>B=1.

tui rất thích lượng giác:

a) = s² + 2s.c +c² +s²- 2s.c + c² =1+1=2

b) = s.c(s/c + c/s) = s.c(s² + c²) / s.c = 1

.............................bài nào cx dễ

( k có việc j khó, chỉ sợ lòng k bền....)

1) \(\left(\tan\alpha+\cot\alpha\right)^2-\left(\tan\alpha-\cot\alpha\right)^2\)

= \(\tan^2\alpha+\cot^2\alpha+2\tan\alpha.\cot\alpha-\tan^2\alpha+2\tan\alpha.\cot\alpha-\cot^2\alpha\)

= \(4\tan\alpha.\cot\alpha\)

= \(4.\frac{\cos\alpha}{\sin\alpha}.\frac{\sin\alpha}{\cos\alpha}=4\)

2) \(\frac{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2}}}\)

= \(\frac{4-2-\sqrt{2+\sqrt{2}}}{\left(2+\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)

= \(\frac{1}{\left(2+\sqrt{2+\sqrt{2+\sqrt{2}}}\right)}\)

Mặt khác: \(\sqrt{2}< 2\Rightarrow2+\sqrt{2}< 4\Rightarrow2+\sqrt{2+\sqrt{2}}< 2+\sqrt{4}=4\)

=> \(2+\sqrt{2+\sqrt{2+\sqrt{2}}}< 2+\sqrt{4}=4\)

=> \(\frac{1}{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}>\frac{1}{4}\)

=> \(\frac{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2}}}>\frac{1}{4}\)

\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)

\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)

a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)

b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)

Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)

\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)

Mình viết luôn là sin với cos, bạn tự cho thêm \(\alpha\) nhé.

VT= \(\sin^2.\dfrac{\sin}{\cos}+\cos^2.\dfrac{\cos}{\sin}+2\sin\cos\)

= \(\dfrac{\sin^3}{\cos}+\dfrac{\cos^3}{\sin}+2\sin\cos\)

= \(\dfrac{\sin^4+\cos^4+2\sin^2.\cos^2}{\cos.\sin}\)

= \(\dfrac{\left(\sin^2+\cos^2\right)^2}{\cos.\sin}\)

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

VP = \(\dfrac{\sin}{\cos}+\dfrac{\cos}{\sin}\)

= \(\dfrac{\sin^2+\cos^2}{\cos.\sin}\)

= \(\dfrac{1}{\cos.\sin}\)(2)

từ (1) và (2) => VT=VP (đpcm)

Chúc bạn học tốt!

cảm ơn bạn

Bạn ơi bài này của lớp 9 chứ có phải của lớp 8 đâu 

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

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

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

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

\(=1-sin^2a+sin^2a=1\)