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tan a=2
=>sin a=2*cosa
\(P=\dfrac{10cosa-3cosa}{cosa+2\cdot2cosa}=\dfrac{7}{5}\)
\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)
\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)
Nhân cả tử và mẫu của phân số chứa tan với \(sina.cosa\)
\(A=\frac{sin^2x-cos^2x}{sin^2x+cos^2x}+cos2x=sin^2x-cos^2x+cos2x=-cos2x+cos2x=0\)
\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}=\frac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin4a.cos4a+2cos^22a-1}\)
\(B=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(C=\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a-1\right)}{2\left(cos^22a+2cos2a+1\right)}\)
\(C=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(D=\frac{sin^22a+4sin^4a-\left(2sina.cosa\right)^2}{4-4sin^2a-sin^22a}=\frac{sin^22a+4sin^4a-sin^22a}{4\left(1-sin^2a\right)-\left(2sina.cosa\right)^2}=\frac{4sin^4a}{4cos^2a-4sin^2a.cos^2a}\)
\(=\frac{sin^4a}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^2a.cos^2a}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).
Ta có:
\(\dfrac{cot\alpha-tan\alpha}{cot\alpha+tan\alpha}=\dfrac{cot\alpha.cot\alpha-cot\alpha tan\alpha}{cot\alpha.cot\alpha+cot\alpha tan\alpha}=\dfrac{cot^2\alpha-1}{cot^2\alpha+1}\)
\(=\dfrac{\dfrac{1}{sin^2\alpha}-2}{\dfrac{1}{sin^2\alpha}}=1-2sin^2\alpha=1-2\left(\dfrac{2}{3}\right)^2=\dfrac{1}{9}\).
\(A=\frac{\frac{sina}{cos^3a}-\frac{cosa}{cos^3a}}{tan^3a+3+\frac{2sina}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}-\frac{1}{cos^2a}}{tan^3a+3+2tana.\frac{1}{cos^2a}}\)
\(=\frac{tana\left(1+tan^2a\right)-\left(1+tan^2a\right)}{tan^3a+3+2tana\left(1+tan^2a\right)}=\frac{3\left(1+9\right)-\left(1+9\right)}{27+3+2.3.\left(1+9\right)}=...\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
Câu 1:
\(sina+cosa=\frac{\sqrt{2}}{2}\Leftrightarrow\left(sina+cosa\right)^2=\frac{1}{2}\)
Chia 2 vế cho \(cos^2a:\) :
\(\left(\frac{sina+cosa}{cosa}\right)^2=\frac{1}{2}.\frac{1}{cos^2a}\Leftrightarrow\left(tana+1\right)^2=\frac{1}{2}\left(1+tan^2a\right)\)
\(\Leftrightarrow tan^2a+4tana+1=0\)
Tiếp tục chia 2 vế cho \(tana\): :
\(\Rightarrow tana+4+cota=0\Rightarrow tana+cota=-4\)
\(P=tan^2a+cot^2a=tan^2a+2+cot^2a-2=\left(tana+cota\right)^2-2=\left(-4\right)^2-2=14\)
Câu 2:
\(3cosa+2sina=2\Rightarrow cosa=\frac{2-2sina}{3}=\frac{2}{3}\left(1-sina\right)\)
Mặt khác ta luôn có: \(sin^2a+cos^2a=1\Leftrightarrow sin^2a+\frac{4}{9}\left(1-sina\right)^2=1\)
\(\Leftrightarrow9sin^2a+4sin^2a-8sina+4=9\)
\(\Leftrightarrow13sin^2a-8sina-5=0\Rightarrow\left[{}\begin{matrix}sina=1>0\left(l\right)\\sina=-\frac{5}{13}\end{matrix}\right.\)
Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)
A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)