Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)
b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)
Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
tan a=2
=>sin a=2*cosa
\(P=\dfrac{10cosa-3cosa}{cosa+2\cdot2cosa}=\dfrac{7}{5}\)
cotα = \(\frac{1}{3}\) \(\Leftrightarrow\frac{cos\alpha}{\sin\alpha}=\frac{1}{3}\Leftrightarrow\sin\alpha=3\cos\alpha\)
cotα =\(\frac{1}{\tan\alpha}=\frac{1}{3}\Rightarrow\tan\alpha=3\)
T = \(\frac{2016}{\sin^2\alpha-\sin\alpha\cos\alpha-\cos^2\alpha}=\frac{2016}{9\cos^2\alpha-3\cos^2\alpha-\cos^2\alpha}\) \(=\frac{2016}{5\cos^2\alpha}=\frac{2016}{5}\times\frac{1}{\cos^2\alpha}=\frac{2016}{5}\times\left(1+\tan^2\alpha\right)\) \(=\frac{2016}{5}\left(1+9\right)=4032\)
Do \(90< a< 180\Rightarrow cosa< 0\Rightarrow tana< 0\Rightarrow\) đề bài sai do tana không thể bằng 3
Nhưng kệ cứ tính thì:
Chia cả tử và mẫu của A cho \(cos^3a\) và lưu ý \(\frac{1}{cos^2a}=1+tan^2a\)
\(A=\frac{tana.\frac{1}{cos^2a}+tan^2a+1}{tan^3a-tana-1}=\frac{tana\left(1+tan^2a\right)+tan^2a+1}{tan^3a-tana-1}\)
Tới đây thay số vào và bấm máy là xong
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)
A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)
OK
\(A=\frac{\frac{sina}{cos^3a}-\frac{cosa}{cos^3a}}{tan^3a+3+\frac{2sina}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}-\frac{1}{cos^2a}}{tan^3a+3+2tana.\frac{1}{cos^2a}}\)
\(=\frac{tana\left(1+tan^2a\right)-\left(1+tan^2a\right)}{tan^3a+3+2tana\left(1+tan^2a\right)}=\frac{3\left(1+9\right)-\left(1+9\right)}{27+3+2.3.\left(1+9\right)}=...\)