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\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)
\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)
a/ Ta có: \(tan\alpha=5\Rightarrow cot\alpha=\frac{1}{5}\) . Đề: \(\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}=\frac{\frac{1}{sin^2\alpha}}{1+\frac{cos^3\alpha}{sin^3\alpha}}=\frac{1+cot^2\alpha}{1+cot^3\alpha}=\frac{1+\left(\frac{1}{5}\right)^2}{1+\left(\frac{1}{5}\right)^3}=\frac{65}{63}\)
b/ Ta có vế trái \(=\frac{sin^2x+cos^2x+cos^2x-sin^2x+\left(sinx+sin3x\right)}{1+2sinx}=\frac{2cos^2x+2.sin2x.cosx}{1+2sinx}=\frac{2cos^2x+4.sinx.cos^2x}{1+2sinx}=\frac{2cos^2x.\left(1+2sinx\right)}{1+2sinx}=2cos^2x\) ( = vế phải)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
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 = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)
\(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}=\dfrac{\dfrac{4sin\alpha}{cos\alpha}+\dfrac{5cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{4tan\alpha+5}{2tan\alpha-3}\)
Biết \(tan\)=\(\dfrac{1}{3}\) nên ta có:
\(\dfrac{4\times\dfrac{1}{2}+5}{2\times\dfrac{1}{2}-3}=\dfrac{2+5}{2-3}=\dfrac{7}{-2}=\dfrac{-7}{2}\)
\(1+tan^2\frac{a}{2}=\frac{1}{cos^2\frac{a}{2}}\Rightarrow cos^2\frac{a}{2}=\frac{1}{1+tan^2\frac{a}{2}}=\frac{1}{5}\)
\(A=\frac{1+5\left(2cos^2\frac{a}{2}-1\right)}{3-2\left(2cos^2\frac{a}{2}-1\right)}=\frac{-4+10cos^2\frac{a}{2}}{5-4cos^2\frac{a}{2}}=\frac{-4+10.\frac{1}{5}}{5-4.\frac{1}{5}}=...\)