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\(tana=\sqrt{3}\)
=>\(\dfrac{sina}{cosa}=\sqrt{3}\)
=>\(sina=\sqrt{3}\cdot cosa\)
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+3=4\)
=>\(cos^2a=\dfrac{1}{4}\)
=>\(cosa=\dfrac{1}{2}\)
=>\(sina=\dfrac{\sqrt{3}}{2}\)
\(A=\dfrac{sin^2a-cos^2a}{sina\cdot cosa}\)
\(=\dfrac{\dfrac{3}{4}-\dfrac{1}{4}}{\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}}=\dfrac{2}{4}:\dfrac{\sqrt{3}}{4}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\)
\(\left(3+\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}+\sqrt{5}\right)-\left(3-\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}-\sqrt{5}\right)=\sqrt{34.64911064}\)
\(A=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
C1:A2=\(3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
A2=\(6+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
A2=\(6+2\sqrt{9-5}\)
A2=6+4=10
A=\(\sqrt{10}\)
\(A=\sqrt{\left(3+\sqrt{5}\right)}+\sqrt{\left(3-\sqrt{5}\right)}\)
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right).\left(3-\sqrt{5}\right)}\)
\(A^2=6+2\sqrt{9-5}\)
\(A^2=6+2\sqrt{4}\)
\(A^2=8\)
\(\Rightarrow A=\sqrt{8}\)
d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)
a) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|+\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)
\(=\sqrt{5}+\sqrt{5}\)
\(=2\sqrt{5}\)
b) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=\left|\sqrt{2}-1\right|-\left|\sqrt{2}-5\right|\)
\(=\sqrt{2}-1-\left(5-\sqrt{2}\right)\)
\(=\sqrt{2}-1-5+\sqrt{2}\)
\(=2\sqrt{2}-6\)
d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)
\(270^0< a< 360^0\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{5}}{3}\)