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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/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
tuỳ ý giá trị biểu thức
Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)
Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
+) ta có : \(A=\left(tan\alpha+cot\alpha\right)^2-\left(tan\alpha-cot\alpha\right)^2\)
\(=tan^2\alpha+cot^2\alpha+2-tan^2\alpha-cot^2\alpha+2=4\) (không phụ thuộc vào \(\alpha\)) \(\Rightarrow\) (đpcm)
+) ta có : \(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=\left(\left(sin^2\alpha+cos^2\alpha\right)^2-3sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=1\) (không phụ thuộc vào \(\alpha\) ) \(\Rightarrow\) (đpcm)
1) \(\left(\tan\alpha+\cot\alpha\right)^2-\left(\tan\alpha-\cot\alpha\right)^2\)
= \(\tan^2\alpha+\cot^2\alpha+2\tan\alpha.\cot\alpha-\tan^2\alpha+2\tan\alpha.\cot\alpha-\cot^2\alpha\)
= \(4\tan\alpha.\cot\alpha\)
= \(4.\frac{\cos\alpha}{\sin\alpha}.\frac{\sin\alpha}{\cos\alpha}=4\)
2) \(\frac{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2}}}\)
= \(\frac{4-2-\sqrt{2+\sqrt{2}}}{\left(2+\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)
= \(\frac{1}{\left(2+\sqrt{2+\sqrt{2+\sqrt{2}}}\right)}\)
Mặt khác: \(\sqrt{2}< 2\Rightarrow2+\sqrt{2}< 4\Rightarrow2+\sqrt{2+\sqrt{2}}< 2+\sqrt{4}=4\)
=> \(2+\sqrt{2+\sqrt{2+\sqrt{2}}}< 2+\sqrt{4}=4\)
=> \(\frac{1}{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}>\frac{1}{4}\)
=> \(\frac{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2}}}>\frac{1}{4}\)
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)
Lúc đó bận nên làm tắt :v.
Áp dụng công thức: \(cot\alpha=\frac{cos\alpha}{sin\alpha}\) ta có:
\(GTBT=\frac{\left(\frac{cos\alpha}{sin\alpha}\right)^2-cos^2\alpha}{\left(\frac{cos\alpha}{sin\alpha}\right)^2}+\frac{sin\alpha cos\alpha}{\frac{cos\alpha}{sin\alpha}}=\frac{\frac{1}{sin^2\alpha}-1}{\frac{1}{sin^2\alpha}}+sin^2\alpha=\left(1-sin^2\alpha\right)+sin^2\alpha=1\)
Là sao ạ :V