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\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)
\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)
\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)
\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
Vì \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\), lại có \(\tan\alpha=\frac{1}{3}\)\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{1}{3}\)\(\Rightarrow\cos\alpha=3\sin\alpha\)
Mà \(\sin^2\alpha+\cos^2\alpha=1\)\(\Rightarrow\sin^2\alpha+9\sin^2\alpha=1\)\(\Rightarrow10\sin^2\alpha=1\)\(\Rightarrow\sin^2\alpha=\frac{1}{10}\)\(\Rightarrow\sin\alpha=\frac{\sqrt{10}}{10}\)
Ta có \(A=\frac{1-2\sin\alpha\cos\alpha}{\sin^2\alpha-\cos\alpha}=\frac{1-2\sin\alpha.3\sin\alpha}{\sin^2\alpha-3\sin\alpha}=\frac{1-6\sin^2\alpha}{\sin^2\alpha-3\sin\alpha}=\frac{1-\frac{6}{10}}{\frac{1}{10}-\frac{3\sqrt{10}}{10}}=\frac{\frac{4}{10}}{\frac{1-3\sqrt{10}}{10}}\)
\(=\frac{4}{1-3\sqrt{10}}=\frac{4\left(1+3\sqrt{10}\right)}{\left(1-3\sqrt{10}\right)\left(1+3\sqrt{10}\right)}=\frac{4\left(1+3\sqrt{10}\right)}{1^2-\left(3\sqrt{10}\right)^2}=\frac{4\left(1+3\sqrt{10}\right)}{1-90}=\frac{4+12\sqrt{10}}{-89}\)
Đề sai nhé
Gía trị của \(1-2\sin.a.\cos\)không thể có giá trị \(\tan a=\frac{1}{3}\)
HT