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NV
28 tháng 11 2019

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

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

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

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

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

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

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

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

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

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

31 tháng 7 2019

cái câu 1 kia lạ thật, phần phía trc có ngoặc thì phải nhân vs hạng tử nào đó chứ nhỉ? Và mk tính ra kq là \(-\cos^22\alpha\)

\(VT=\cos^4\alpha+\sin^4\alpha-2\cos^6\alpha-2\sin^6\alpha\)

\(=\sin^4\alpha\left(1-2\sin^2\alpha\right)-\cos^4\alpha\left(2\cos^2\alpha-1\right)\)

\(=\sin^4\alpha.\cos2\alpha-\cos^4\alpha.\cos2\alpha\)

\(=\cos2\alpha\left(\sin^2\alpha.\sin^2\alpha-\cos^4\alpha\right)\)

\(=\cos2\alpha.\left[\left(1-\cos^2\alpha\right)^2-\cos^4\alpha\right]\)

\(=\cos2\alpha.\left(1-2\cos^2\alpha\right)\)

\(=-\cos^22\alpha\)

2/ \(VT=\frac{1-\cos^2\alpha+\cos^2\alpha}{1+\sin2\alpha}=\frac{1}{1+\sin2\alpha}\)

\(VP=\frac{\frac{\sin\alpha}{\cos\alpha}-1}{\frac{\sin\alpha}{\cos\alpha}+1}=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)

hmm, câu 2 có vẻ vô lí, bn thử nhân chéo lên mà xem, nó ko ra KQ = nhau đâu

AH
Akai Haruma
Giáo viên
31 tháng 7 2019

1)

\((\cos^4a+\sin ^4a)-2(\cos^6a+\sin ^6a)=(\cos ^4a+\sin ^4a)-2(\cos ^2a+\sin ^2a)(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=(\cos ^4a+\sin ^4a)-2(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=-(\cos ^4a-2\sin ^2a\cos ^2a+\sin ^4a)=-(\cos ^2a-\sin ^2a)^2=-\cos ^22a\)

(bạn xem lại đề. Nếu thay $(\cos ^4a+\sin ^4a)$ thành $3(\cos ^4a+\sin ^4a)$ thì kết quả thu được là $(\cos ^2a+\sin ^2a)^2=1$ như yêu cầu)

2) Sửa đề:

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

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

Bạn lưu ý viết đề bài chuẩn hơn.

AH
Akai Haruma
Giáo viên
30 tháng 4 2019

Lời giải:

a)

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

b)

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

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

c)

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

\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

AH
Akai Haruma
Giáo viên
2 tháng 5 2019

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.

9 tháng 10 2016

\(sina+cosa=\sqrt{2}\Leftrightarrow\left(sina+cosa\right)^2=2\\ \)

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

\(\Leftrightarrow1+2.sina.cosa=2\)

\(\Leftrightarrow2.sina.cosa=2-1=1\)

\(\Leftrightarrow\sin a.cosa=\frac{1}{2}\)

Vậy  P=sina.cosa=\(\frac{1}{2}\)

\(Q=\sin^4a+cos^4a\)

\(\Leftrightarrow\left(sin^2a\right)^2+\left(cos^2a\right)^2\)

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

\(\Leftrightarrow1^2-2.sin^2a.cos^2a\) tách tiếp rồi thế vào là được .tương tự phàn P ý
còn R thì tách sin^3a=sin^2a+sina tương tự cos mũ 3 a cụng vậy
theo tớ là như thế còn có sai thì đừng có ném đá ném gạch na

 

 

NV
1 tháng 5 2019

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

\(=\frac{1}{sin^2a}\left(tan^3a-tana+cota\right)=\left(1+cot^2a\right)\left(tan^3a-tana+cota\right)\)

\(=tan^3a-tana+cota+cot^2a.tan^3a-cot^2a.tana+cot^3a\)

\(=tan^3a-tana+cota+tana-cota+cot^3a\)

\(=tan^3a+cot^3a\)

1 tháng 5 2019

Em cảm ơn

NV
3 tháng 6 2020

\(\frac{sinA}{cosA}+\frac{sinB}{cosB}=\frac{2cos\frac{C}{2}}{sin\frac{C}{2}}\Leftrightarrow\frac{sinA.cosB+cosA.sinB}{cosA.cosB}=\frac{2sin\frac{C}{2}.cos\frac{C}{2}}{sin^2\frac{C}{2}}\)

\(\Leftrightarrow\frac{sin\left(A+B\right)}{cosA.cosB}=\frac{2sinC}{1-cosC}\Leftrightarrow\frac{sinC}{cosA.cosB}=\frac{2sinC}{1-cosC}\)

\(\Leftrightarrow1-cosC=2cosA.cosB=cos\left(A+B\right)+cos\left(A-B\right)\)

\(\Leftrightarrow1-cosC=-cosC+cos\left(A-B\right)\)

\(\Leftrightarrow cos\left(A-B\right)=1\Rightarrow A-B=0\Rightarrow A=B\)

\(\Rightarrow\) Tam giác ABC cân tại C

\(\frac{cos^2A+cos^2B}{sin^2A+sin^2B}=\frac{1}{2}\left(cot^2A+cot^2B\right)\)

\(\Leftrightarrow2cos^2A+2cos^2B=\left(sin^2A+sin^2B\right)\left(cot^2A+cot^2B\right)\)

\(\Leftrightarrow2cos^2A+2cos^2B=cos^2A+cos^2B+sin^2A.cot^2B+sin^2B.cot^2A\)

\(\Leftrightarrow cos^2A+cos^2B=\frac{sin^2A.cos^2B}{sin^2B}+\frac{sin^2B.cos^2A}{sin^2A}\)

\(\Leftrightarrow cos^2A\left(\frac{sin^2B}{sin^2A}-1\right)=cos^2B\left(1-\frac{sin^2A}{sin^2B}\right)\)

\(\Leftrightarrow\frac{cos^2A\left(sin^2B-sin^2A\right)}{sin^2A}=\frac{cos^2B\left(sin^2B-sin^2A\right)}{sin^2B}\)

\(\Leftrightarrow cot^2A\left(sin^2B-sin^2A\right)=cot^2B\left(sin^2B-sin^2A\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2B=sin^2A\\cot^2A=cot^2B\end{matrix}\right.\) \(\Rightarrow A=B\)

NV
8 tháng 6 2020

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

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

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

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

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

AH
Akai Haruma
Giáo viên
28 tháng 4 2019

Lời giải:

a)

\(\frac{\cos (a-b)}{\cos (a+b)}=\frac{\cos a\cos b+\sin a\sin b}{\cos a\cos b-\sin a\sin b}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\)

b)

\(2(\sin ^6a+\cos ^6a)+1=2(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)

\(=2(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)

\(=3(\sin ^4a+\cos ^4a)-(\sin ^4a+\cos ^4a+2\sin ^2a\cos ^2a)+1\)

\(=3(\sin ^4a+\cos ^4a)-(\sin ^2a+\cos ^2a)^2+1\)

\(=3(\sin ^4a+\cos ^4a)-1^2+1=3(\sin ^4a+\cos ^4a)\)

c)

\(\frac{\tan a-\tan b}{cot b-\cot a}=\frac{\tan a-\tan b}{\frac{1}{\tan b}-\frac{1}{\tan a}}\) (nhớ rằng \(\tan x.\cot x=1\rightarrow \cot x=\frac{1}{\tan x}\) )

\(=\frac{\tan a-\tan b}{\frac{\tan a-\tan b}{\tan a\tan b}}=\tan a\tan b\)

d)

\((\cot x+\tan x)^2-(\cot x-\tan x)^2=(\cot ^2x+\tan ^2x+2\cot x\tan x)-(\cot ^2x-2\cot x\tan x+\tan ^2x)\)

\(=4\cot x\tan x=4.1=4\)

e)

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

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

Vậy ta có đpcm.