Chứng minh đẳng thức nhé các bạn !!! Mình quên ghi đầu bài

a: \(\sin^2a+\cos^2a=1\)

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

hay \(\dfrac{\cos a}{1-\sin a}=\dfrac{1+\sin a}{\cos a}\)

b: \(VT=\dfrac{\left(\sin a+\cos a+\sin a-\cos a\right)\left(\sin a+\cos a-\sin a+\cos a\right)}{\sin a\cdot\cos a}\)

\(=\dfrac{2\cdot\cos a\cdot2\sin a}{\sin a\cdot\cos a}=4\)

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

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

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

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

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

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

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

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

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

\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

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

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

\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)

b) khai triển hằng đẳng thức là ra

a) nhân tích chéo

\(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)\(\Leftrightarrow\cos^2\alpha+\sin^2\alpha=1\)(luôn đúng)

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

\(=\frac{4\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}=4\)(đpcm)

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

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

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

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

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

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

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

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

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

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

Lời giải:

\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)

\(=(1+\frac{\cos a}{\sin a})\sin ^3a+(1+\frac{\sin a}{\cos a})\cos ^3a\)

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

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

Sao ko thấy tham số m nào bạn nhỉ?

\(VT=\dfrac{1}{2}sin2a-tan^2a.cota+\dfrac{1}{2}tan^2a.sin2a\)

\(=\dfrac{1}{2}sin2a\left(1+tan^2a\right)-tana\)

\(=\dfrac{1}{2}sin2a.\dfrac{1}{cos^2a}-tana\)

\(=\dfrac{sina.cosa}{cos^2a}-tana=tana-tana=0\) (đpcm)