Question : $\triangle$XYZ is right angled at Y. If $\angle$X = 60°, then find the value of $(\sec Z+\frac{2}{\sqrt3})$.
Option 1: $\frac{4}{\sqrt3}$
Option 2: $\frac{\sqrt2+2}{2\sqrt2}$
Option 3: $\frac{7}{2\sqrt3}$
Option 4: $\frac{4}{2\sqrt3}$
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Correct Answer: $\frac{4}{\sqrt3}$
Solution :
Given: $\triangle$XYZ is right angled at Y and $\angle$X = 60°.
$(\sec Z+\frac{2}{\sqrt3})$
= $(\sec(90°-60°)+\frac{2}{\sqrt3})$
= $(\sec30°+\frac{2}{\sqrt3})$
= $(\frac{2}{\sqrt3}+\frac{2}{\sqrt3})$
= $\frac{4}{\sqrt3}$
Hence, the correct answer is $\frac{4}{\sqrt3}$.
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