`(sin \alpha+cos \alpha)^2-2sin \alpha.cos \alpha`

`=sin^2 \alpha + cos^2 \alpha+2sin \alpha.cos \alpha-2sin \alpha.cos \alpha`

`=1+0`

`=1`.

[1-2sina/2cosa/2+(2cos^2a/2 - 1)]/[1-2sina/2cosa/2-1+2sin^a]

=2cosa/2(cosa/2-sina/2)/[2sina/2(sina/2-cosa/2)]

= -cota/2

Giả sử các biểu thức đều xác định

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

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

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

Chứng minh các hằng đẳng thức trên

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)

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

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

\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)

\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)

\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)

\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)

Lời giải:

1.

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

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

2.

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

\(=\cos a-\sin a\)

3.

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

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

4.

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

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

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

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

Chà, bạn coi lại đề, \(\frac{1-sin^2a}{cosa+sina}\) hay \(\frac{cos^2a-sin^2a}{cosa+sina}\)