Lời giải:
\(A=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a(\sin ^2a+\cos ^2a)\)

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

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

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

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

$=\frac{-4\sin a\cos a}{\sin a\cos a}=-4$

\(B=\left(sina+cosa\right)^2-\left(cosa-sina\right)^2=\left(sin^2a+2sinacosa+cos^2a\right)-\left(cos^2a-2cosasina+sin^2a\right)=sin^2a+2sinacosa+cos^2a-cos^2a+2cosasina-sin^2a=4sinacosa\)\(A=\dfrac{1+2sinacosa}{sina+cosa}=\dfrac{sin^2a+cos^2a+2cosasina}{sina+cosa}=\dfrac{\left(sina+cosa\right)^2}{sina+cosa}=sina+cosa\)

C mik bó tay

a, Sử dụng tích chéo:

Ta có:

+/ \(\cos\alpha.\cos\alpha=\cos^2\alpha\) (1)

+/ \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=1-\sin^2\alpha\)

Mà \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow1-\sin^2\alpha=\cos^2\alpha\)

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

Từ (1), (2)

\(\Rightarrow\)\(\cos\alpha.\cos\alpha=\)\(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)\)

\(\Rightarrow\)\(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\) (đpcm)

b/ xem lại đề

Bạn không ghi rõ yêu cầu đề bài thì làm sao mà làm?

Lời giải:
a)

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

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

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

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

b)

\(2(\sin a-\cos a)^2-[(\sin a+\cos a)^2+\sin a\cos a]\)

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

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

\(=1-7\sin a\cos a\)

c)

\((\tan a-\cot a)^2-(\tan a+\cot a)^2\)

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

\(=-4\tan a\cot a=-4\)