tan a =2/3

=> đặt sin a = 2x thì cos a = 3x

rồi làm tiếp còn cách khác thì k biết làm

Ta có: \(tan\alpha=3=\frac{sin\alpha}{cos\alpha}\Rightarrow sin\alpha=3cos\alpha\)

Suy ra: \(B=\frac{\left(sin\alpha-cos\alpha\right)\left(sin^2\alpha+cos^2\alpha+sin\alpha.cos\alpha\right)}{\left(sin\alpha+cos\alpha\right)\left(sin^2\alpha+cos^2\alpha-sin\alpha.cos\alpha\right)}\)

\(=\frac{2cos\alpha.\left(1+3cos^2\alpha\right)}{4cos\alpha.\left(1-3cos^2\alpha\right)}=\frac{1+3cos^2\alpha}{2.\left(1-3cos^2\alpha\right)}\)

khó quá chị ơi

tan.a ?

\(sin^2a-sina.cosa+cos^2a\)

\(\Leftrightarrow tan^2a-tana+1\)

Thay tana = 1/2

\(\left(\frac{1}{2}\right)^2-\frac{1}{2}+1=\frac{3}{4}\)

VT = sin3a.cos^3a + sin^3a.cos3a 
= sin3a.cosa.cos^2a + sin^2a.sina.cos3a 
= 1/2.(sin2a + sin4a).cos^2a + 1/2.sin^2a.(sin(-2a) + sin4a) 
= 1/2.(sin2a + sin4a).cos^2a + 1/2.sin^2a.(sin4a - sin2a) 
= 1/2.sin2a.cos^2a + 1/2.sin4a.cos^2a + 1/2.sin^2a.sin4a - 1/2.sin^2a.sin2a 
= 1/2.sin2a.(cos^2a - sin^2a) + 1/2.sin4a.(cos^2a + sin^2a) 
= 1/2.sin2a.cos2a + 1/2.sin4a 
= 1/4.sin4a + 1/2.sin4a 
= 3/4.sin4a = VP 
=> đpcm

P/s: Chỉ sợ you ko hiểu

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

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

\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

\(=1-3sin^2a.cos^2a\)

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

\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này

\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)

\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)

\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)

\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)

\(D=\frac{\frac{sina}{cos^3a}+\frac{5cosa}{cos^3a}}{\frac{sin^3a}{cos^3a}-\frac{2cos^3a}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}+\frac{5}{cos^2a}}{tan^3a-2}=\frac{tana\left(1+tan^2a\right)+5\left(1+tan^2a\right)}{tan^3a-2}\)

Bạn thay số và bấm máy

Lời giải:

Ta có:

\(\frac{\tan ^3a}{\sin ^2a}-\frac{1}{\sin a\cos a}+\frac{\cot ^3a}{\cos ^2a}=\frac{\tan ^3a\cos ^2a+\cot ^3a\sin ^2a}{\sin ^2a\cos ^2a}-\frac{\sin a\cos a}{\sin ^2a\cos ^2a}\)

\(=\frac{\frac{\sin ^3a}{\cos ^3a}.\cos ^2a+\frac{\cos ^3a}{\sin ^3a}.\sin ^2a}{\sin ^2a\cos ^2a}-\frac{\sin a\cos a}{\sin ^2a\cos ^2a}\)

\(=\frac{\frac{\sin ^3a}{\cos a}+\frac{\cos ^3a}{\sin a}-\sin a\cos a}{\sin ^2a\cos ^2a}=\frac{\sin ^4a+\cos ^4a-\sin ^2a\cos ^2a}{\sin ^3a\cos ^3a}\)

\(=\frac{(\sin ^2a+\cos ^2a)(\sin ^4a+\cos ^4a-\sin ^2a\cos ^2a)}{\sin ^3a\cos ^3a}\)

\(=\frac{\sin ^6a+\cos ^6a}{\sin ^3a\cos ^3a}=\frac{\sin ^3a}{\cos ^3a}+\frac{\cos ^3a}{\sin ^3a}=\tan ^3a+\cot ^3a\)

Ta có đpcm.

\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)

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

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

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

\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)

\(\frac{sin^3a+cos^3a}{sina+cosa}=\frac{\left(sina+cosa\right)\left(sin^2a+cos^2a-sina.cosa\right)}{sina+cosa}\)

\(=sin^2a+cos^2a-sina.cosa\)

\(=1-sina.cosa\)