1easy
:
let
\frac{\sqrt{3}}{2}=\frac{3}{t_1}-\frac{4}{t_1^3} \\ \sqrt{3}t_1 =6-\frac{8}{t_1^2}\\ also \\ \\ \frac{1}{2}=\frac{4}{t_2^3}-\frac{3}{t_2} \\ t_2=\frac{8}{t_2^2}-6 \\ \sqrt{3}t_1-t_2=12-8\left(\frac{1}{t_1^2}+\frac{1}{t_2^2} \right)=4
Find the value of:
\sqrt 3 \csc 20^{\circ} - \sec 20^{\circ}
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1 Answers
1