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

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

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

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

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

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

phần chứng minh biểu thức không phụ thuộc \(x\)

ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)

\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)

ý còn lại : xem lại đề nha bn

phần chứng minh đẳng thức

ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)

\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)

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

\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)

ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)

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

\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)

còn 2 câu kia để chừng nào rảnh mk giải cho nha

mk lm 2 câu còn lại nha

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

\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)

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

\(=sinx+cosx\left(đpcm\right)\)

ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)

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

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(\left(sina.cosb-cosa.sinb\right)\left(sina.cosb+cosa.sinb\right)\right)}\)

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-sin\left(a-b\right)sin\left(a+b\right)}=-cos^2a.sin^2b\left(đpcm\right)\)

mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm

Xét ΔBAC vuông tại B có a = ^A ta có :

a) \(\frac{\sin\alpha}{\cos\alpha}=\frac{\sin A}{\cos A}=\frac{\frac{BC}{AB}}{\frac{AB}{AC}}=\frac{BC}{AB}\cdot\frac{AC}{AB}=\frac{BC}{AB}=\tan A=\tan\alpha\left(đpcm\right)\)

b) \(\frac{\cos\alpha}{\sin\alpha}=\frac{\cos A}{\sin A}=\frac{\frac{AB}{AC}}{\frac{BC}{AC}}=\frac{AB}{AC}\cdot\frac{AC}{BC}=\frac{AB}{BC}=\cot A=\cot\alpha\left(đpcm\right)\)

c) \(\tan\alpha\cdot\cot\alpha=\tan A\cdot\cot A=\frac{BC}{AB}\cdot\frac{AB}{BC}=1\left(đpcm\right)\)

d) \(\sin^2\alpha+\cos^2\alpha=\sin^2A+\cos^2A=\frac{BC^2}{AC^2}+\frac{AB^2}{AC^2}=\frac{AB^2+BC^2}{AC^2}=1\left(đpcm\right)\)

e) \(\frac{1}{\cos^2\alpha}=\frac{1}{\cos^2A}=\frac{1}{\frac{AB^2}{AC^2}}=\frac{AC^2}{AB^2};1+\tan^2\alpha=1+\tan^2A=1+\frac{BC^2}{AB^2}=\frac{AB^2+BC^2}{AB^2}=\frac{AC^2}{AB^2}\)

\(\Rightarrow1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\left(đpcm\right)\)

f) \(\frac{1}{\sin^2\alpha}=\frac{1}{\sin^2A}=\frac{1}{\frac{BC^2}{AC^2}}=\frac{AC^2}{BC^2};1+\cot^2\alpha=1+\cot^2A=1+\frac{AB^2}{BC^2}=\frac{BC^2+AB^2}{BC^2}=\frac{AC^2}{BC^2}\)

\(\Rightarrow1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\left(đpcm\right)\)

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

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

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

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

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

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

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

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

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

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

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

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.

3)

\(\sin ^4a+\cos ^4a-\sin ^6a-\cos ^6a=\sin ^4a+\cos ^4a-[(\sin ^2a)^3+(\cos ^2a)^3]\)

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

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

\(=\sin ^2a\cos ^2a\) (đpcm)

4)

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

5)

\(\frac{\tan a}{1-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}=\frac{\tan a}{(tan a\cot a)^2-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}\)

\(=\frac{\tan a}{\tan ^2a(\cot ^2a-1)}.\frac{\cot ^2a-1}{\cot a}=\frac{1}{\tan a\cot a}=\frac{1}{1}=1\)

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Mấu chốt của các bài này là bạn sử dụng 2 công thức sau:

1. \(\sin ^2x+\cos^2x=1\)

2. \(\tan x.\cot x=1\)

Bài 2: 

\(\cos\alpha=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)

\(\tan\alpha=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

\(\cot\alpha=\dfrac{\sqrt{5}}{2}\)