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1-sin60* = tan60*-1
Given: 1-sin60* = tan60*-1
To do: Prove LHS =RHS
Solution:
Let us simplify LEFT HAND SIDE:
= $\frac{1 -sin60°}{cos60°}$
= $\frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}}$
= $(2-\sqrt{3})$
Let us simplify Right HAND SIDE
= $\frac{ tan60° -1}{tan60° +1}$
=$\frac{\sqrt{3} -1}{\sqrt{3} +1}$
=$\frac{(\sqrt{3} -1)(\sqrt{3} -1)}{(\sqrt{3} +1)(\sqrt{3} -1)}$
=$\frac{ \sqrt{3^2} +1^2 -2√3}{\sqrt{3^2}-1}$
=$\frac{ 4-2\sqrt{3}}{2}$
= $2 -\sqrt{3}$
Therefore, LHS = RHS
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