Lời giải:

a)

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

\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)

b)

Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:

\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)

\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)

\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)

\(=1+0+1=2\)

Lời giải:

a)

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

\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)

b)

Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:

\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)

\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)

\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)

\(=1+0+1=2\)

b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)

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

=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm

đặt \(\sin\alpha=a;\cos\alpha=b\)

khi đó:

\(a+b=\frac{7}{5}\Leftrightarrow a^2+b^2+2ab=\frac{49}{25}\)

\(\Leftrightarrow1+2ab=\frac{49}{25}\Leftrightarrow2ab=\frac{24}{25}\Leftrightarrow ab=\frac{12}{25}\)

ta có

\(\left\{{}\begin{matrix}a+b=\frac{7}{5}\\ab=\frac{12}{25}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left(\frac{7}{5}-b\right)b=\frac{12}{25}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\b^2-\frac{7}{5}b+\frac{12}{25}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left(b-\frac{3}{5}\right)\left(b-\frac{4}{5}\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left[{}\begin{matrix}b=\frac{3}{5}\\b=\frac{4}{5}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\frac{3}{5}\\b=\frac{4}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\frac{4}{5}\\b=\frac{3}{5}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{3}{4}\\\frac{a}{b}=\frac{4}{3}\end{matrix}\right.\)\(\)

hay tan \(\alpha\approx37^o\)hoặc tan\(\alpha\approx53^o\)

câu 1 : ta có : \(A=\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-\left(sin^8x+cos^8x\right)\)

\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-3sin^2x.cos^2x\right)\)

\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-sin^2x.cos^2x\right)+2sin^2xcos^2x\)

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

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

tới đây mk xin sử dụng kiến thức lớp 10 một chút

\(=\dfrac{sin^22x}{4}\left(1+\dfrac{sin^22x}{4}\right)=\dfrac{sin^22x}{4}+\dfrac{sin^42x}{16}\)

vẩn phụ thuộc vào x \(\Rightarrow\) đề sai .

câu 1 : câu này bn có thể tìm trong trang của mk , mk nhớ đã làm nó rồi nhưng tìm hoài không đc . nếu đc bn có thể chờ mk đi hok về mk sẽ kiếm cho bn hoắc có thể là lm lại cho bn nha :)

câu 2 : https://hoc24.vn/hoi-dap/question/657072.html

câu 3 : https://hoc24.vn/hoi-dap/question/657069.html

câu 4 : https://hoc24.vn/hoi-dap/question/656635.html

câu 5 : https://hoc24.vn/hoi-dap/question/657071.html