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\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)
\(=0+2\cdot1-2=0\)
\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)
\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)
\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)
a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
\(\sin15^0=\dfrac{\sqrt{6}-\sqrt{2}}{4}=\cos75^0\)
\(\tan15^0=\cot75^0=2-\sqrt{3}\)
tan x=2
=>\(cotx=\dfrac{1}{tanx}=\dfrac{1}{2}\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+4=5\)
=>\(cos^2x=\dfrac{1}{5}\)
=>\(cosx=\dfrac{1}{\sqrt{5}}\) hoặc \(cosx=-\dfrac{1}{\sqrt{5}}\)
\(tanx=2\)
=>\(\dfrac{sinx}{cosx}=2\)
=>\(sinx=2\cdot cosx\)
TH1: \(cosx=\dfrac{1}{\sqrt{5}}\)
=>\(sinx=\dfrac{2}{\sqrt{5}}\)
\(H=\dfrac{sinx+cotx}{5sinx-5cosx}\)
\(=\dfrac{\dfrac{2}{\sqrt{5}}+\dfrac{1}{2}}{5\left(\dfrac{2}{\sqrt{5}}-\dfrac{1}{\sqrt{5}}\right)}=\dfrac{4+\sqrt{5}}{2\sqrt{5}}:\sqrt{5}=\dfrac{4+\sqrt{5}}{10}\)
TH2: \(cosx=-\dfrac{1}{\sqrt{5}}\)
=>\(sinx=2\cdot cosx=-\dfrac{2}{\sqrt{5}}\)
\(H=\dfrac{sinx+cotx}{5\left(sinx-cosx\right)}\)
\(=\dfrac{\dfrac{-2}{\sqrt{5}}+\dfrac{1}{2}}{5\left(-\dfrac{2}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}\right)}=\dfrac{-4+\sqrt{5}}{2\sqrt{5}}:\left(-\sqrt{5}\right)\)
\(=\dfrac{4-\sqrt{5}}{10}\)
Ta có:
\(1+tan^2x=\dfrac{1}{cos^2x}\)
\(\Leftrightarrow cos^2x=\dfrac{1}{1+tan^2x}\)
\(\Leftrightarrow cos^2x=\dfrac{1}{1+3^2}\)
\(\Leftrightarrow cosx=\sqrt{\dfrac{1}{10}}=\dfrac{\sqrt{10}}{10}\)
Mà: \(tanx=\dfrac{sinx}{cosx}\)
\(\Leftrightarrow sinx=tanx\cdot cosx\)
\(\Leftrightarrow sinx=3\cdot\dfrac{\sqrt{10}}{10}=\dfrac{3\sqrt{10}}{10}\)
Giá trị của A là:
\(A=\dfrac{\dfrac{3\sqrt{10}}{10}+4\cdot\dfrac{\sqrt{10}}{10}}{2\cdot\dfrac{3\sqrt{10}}{10}-\dfrac{\sqrt{10}}{10}}\)
\(A=\dfrac{\dfrac{3\sqrt{10}}{10}+\dfrac{4\sqrt{10}}{10}}{\dfrac{6\sqrt{10}}{10}-\dfrac{\sqrt{10}}{10}}\)
\(A=\dfrac{\dfrac{7\sqrt{10}}{10}}{\dfrac{5\sqrt{10}}{10}}\)
\(A=\dfrac{7}{5}\)
tan=3
=>sin=3*cos
\(A=\dfrac{sin+4cos}{2sin-cos}=\dfrac{3cos+4cos}{6cos-cos}=\dfrac{7}{5}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
\(D=\sin^215+\sin^275-\dfrac{2\cos49}{\sin41}+\tan26.\tan64\)
\(=\sin^215+\cos^215-\dfrac{2\cos49}{\cos49}+1\)
\(=1-2+1=0\)
Học tốt !!