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+) ta có : \(A=\dfrac{sinx-cosx}{2sinx+cosx}=\dfrac{\dfrac{sinx}{sinx}-\dfrac{cosx}{sinx}}{\dfrac{2sinx}{sinx}+\dfrac{cosx}{sinx}}\) \(=\dfrac{1-cotx}{2+cotx}\)
\(=\dfrac{1-3\sqrt{8}}{2+3\sqrt{8}}=\dfrac{-37+9\sqrt{2}}{34}\)
+) ta có : \(A=\dfrac{1+sin^2x}{2+sinx.cosx}=\dfrac{sin^2x+cos^2x+sin^2x}{2sin^2x+2cos^2x+sinx.cosx}\) \(=\dfrac{2sin^2x+cos^2x}{2sin^2x+2cos^2x+sinx.cosx}=\dfrac{\dfrac{2sin^2x}{sin^2x}+\dfrac{cos^2x}{sin^2x}}{\dfrac{2sin^2x}{sin^2x}+\dfrac{2cos^2x}{sin^2x}+\dfrac{sinx.cosx}{sin^2x}}\) \(=\dfrac{2+cot^2x}{2+2cot^2x+cotx}=\dfrac{2+\left(3\sqrt{8}\right)^2}{2+2\left(3\sqrt{8}\right)^2+3\sqrt{8}}=\dfrac{74}{146+3\sqrt{8}}\)
\(cosa.sina=\frac{1}{5}\Rightarrow\frac{cosa.sina}{sin^2a}=\frac{1}{5sin^2a}=\frac{sin^2a+cos^2a}{5sin^2a}\)
\(\Rightarrow\frac{cosa}{sina}=\frac{1}{5}+\frac{1}{5}.\frac{cos^2a}{sin^2a}\)
\(\Rightarrow cota=\frac{1}{5}+\frac{1}{5}cot^2a\)
\(\Rightarrow cot^2a-5cota+1=0\)
\(\Rightarrow cota=\frac{5\pm\sqrt{21}}{2}\)
Câu 2:
\(\frac{cosa}{1-sina}=\frac{cosa\left(1+sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\frac{cosa\left(1+sina\right)}{1-sin^2a}=\frac{cosa\left(1+sina\right)}{cos^2a}=\frac{1+sina}{cosa}\)
b/
\(\frac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}\)
\(=\frac{sin^2a+cos^2a+2sina.cosa-\left(sin^2a+cos^2a-2sina.cosa\right)}{sina.cosa}\)
\(=\frac{4sina.cosa}{sina.cosa}\)
\(=4\)
\(\frac{sin^2a-cos^2a+cos^4a}{cos^2a-sin^2a+sin^4a}=\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^2a-cos^2a.sin^2a}{cos^2a-sin^2a.cos^2a}\)
\(=\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^2a.sin^2a}{cos^2a.cos^2a}=tan^4a\)
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-sin^2a.cos^2a=1-2sin^2a.cos^2a\)
a) ta có : \(\left(1-cosa\right)\left(1+cosa\right)=1-cos^2a=sin^2a\left(đpcm\right)\)
b) ta có : \(1+sin^2a+cos^2a=1+1=2\left(đpcm\right)\)
c) ta có : \(sina-sina.cos^2a=sina\left(1-cos^2a\right)=sina.sin^2a=sin^3a\left(đpcm\right)\)
d) đề thiếu
