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\(tan10^0.tan80^0.tan20^0.tan70^0.tan30.tan60.tan40.tan50\)
\(=tan10.tan\left(90-10\right).tan20.tan\left(90-20\right).tan30.tan\left(90-30\right).tan40.tan\left(90-40\right)\)
\(=tan10.cot10.tan20.cot20.tan30.cot30.tan40.cot40\)
\(=1.1.1.1=1\)
a: \(\tan80^0>\sin80^0>\sin50^0\)
b: \(\tan40^0< \sin40^0< \sin60^0=\cos30^0\)
\(A=\frac{sin80}{cos80}\left(\frac{sin20}{cos20}+\frac{sin140}{cos140}\right)+\frac{sin140.sin20}{cos140.cos20}\)
\(=\frac{sin80}{cos80}\left(\frac{sin20.cos140+cos20.sin140}{cos20.cos140}\right)+\frac{\frac{1}{2}\left(cos120-cos160\right)}{cos20.cos140}\)
\(=\frac{sin80}{cos80}.\frac{sin160}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}\)
\(=\frac{2sin^280}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}=\frac{1-cos160}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}\)
\(=\frac{2-2cos160+cos120-cos160}{2cos20.cos140}=\frac{\frac{3}{2}-3cos160}{cos120+cos160}=\frac{-3\left(-\frac{1}{2}+cos160\right)}{-\frac{1}{2}+cos160}=-3\)
Chọn B.
Ta có
C = ( tan50 . tan 850 ) .( tan 150 tan 750 ) ...tan 450
= ( tan50 .cot 50 ) .( tan 150 cot 150 ) ..tan 450 = 1
( do với 2 góc phụ nhau thì tan góc này bằng cot góc kia)
a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)
áp dụng: A = 2
b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)
áp dụng: B = 0
c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)
áp dụng: C = 1
Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} = - \cos {70^o} = - \sin {20^o}\)
\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)
Ta có: \(\tan {110^o} = - \tan {70^o} = - \cot {20^o};\;\cot {110^o} = - \cot {70^o} = - \tan {20^o}.\)
\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)
Ta có:
\(\begin{array}{l}\cos {30^o} = \sin \left( {{{90}^o} - {{30}^o}} \right) = \sin {60^o} = \frac{{\sqrt 3 }}{2};\\\sin {150^o} = \sin \left( {{{180}^o} - {{150}^o}} \right) = \sin {30^o} = \frac{1}{2};\\\tan {135^o} = - \tan \left( {{{180}^o} - {{135}^o}} \right) = - \tan {45^o} = - 1\end{array}\)
\( \Rightarrow E = 2.\frac{{\sqrt 3 }}{2} + \frac{1}{2} - 1 = \sqrt 3 - \frac{1}{2}.\)
\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)
\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)
\(\Rightarrow P=4\)
\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)
\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)
\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)
\(tan\left(a+b\right)=7;tan\left(a-b\right)=4\)
\(tan2a=tan\left[\left(a+b\right)+\left(a-b\right)\right]=\dfrac{7+4}{1-7.4}=\dfrac{11}{-27}=-\dfrac{11}{27}\)
gợi ý tan 10o = cot 80o
mà tan a . cot a =1
phần còn lại tự làm
chưa hiểu thì hỏi nhé