Answer:
4
Step-by-step explanation:
[tex]sin(\theta)=\frac{opposite}{hypotenuse}=\frac{1}{\sqrt{3} }\\ \\Therefore, opposite= 1, hypotenuse=\sqrt{3}. \ using \ Pythagoras:\\\\Hypotenuse^2=Adjacent^2+Opposite^2\\\\(\sqrt{3})^2 =Adjacent^2+1^2\\\\3=Adjacent^2+1\\\\Adjacent^2=3-1=2\\\\Adjacent=\sqrt{2}[/tex]
[tex]cos(\theta)=\frac{adjacent}{hypotenuse}=\frac{\sqrt{2} }{\sqrt{3}}\\ \\tan(\theta)=\frac{opposite}{adjacent } = \frac{1}{\sqrt{2}} \\\\Therefore:\\\\(tan(\theta)+\frac{1}{cos(\theta)})^2+ (tan(\theta)-\frac{1}{cos(\theta)})^2=( \frac{1}{\sqrt{2}}+\frac{1}{ \frac{\sqrt{2} }{\sqrt{3}}} )^2+( \frac{1}{\sqrt{2}}-\frac{1}{ \frac{\sqrt{2} }{\sqrt{3}}} )^2\\\\=( \frac{1}{\sqrt{2}}+ \frac{\sqrt{3} }{\sqrt{2}} )^2+( \frac{1}{\sqrt{2}}- \frac{\sqrt{3} }{\sqrt{2}} )^2[/tex]
[tex]=(\frac{1+\sqrt{3} }{\sqrt{2} } )^2+(\frac{1-\sqrt{3} }{\sqrt{2} } )^2=\frac{1+2\sqrt{3}+3 }{2} +\frac{1-2\sqrt{3}+3 }{2} =\frac{1+1+2\sqrt{3}-2\sqrt{3}+3+3 }{2} =\frac{8}{2}=4[/tex]
