Question
Prove that $(\sqrt{3}+1)(3-\cot30^\circ)=\tan^360^\circ-2\sin60^\circ.$
✓
Answer
$=\text{R.H.S}=\tan^360^\circ-2\sin60^\circ $
$=\big(\sqrt{3}\big)^3-2\frac {\sqrt{3}}{2}=3\sqrt{3}-\sqrt{3}=2\sqrt {3}$
$\text{L.H.S}=\big(\sqrt{3}+1\big)\big(3-\cot30^ \circ\big)$
$=\big(\sqrt{3}+1\big)\big(3- \sqrt{3}\big)$
$=\big(\sqrt{3}+1\big)\sqrt {3}\big(\sqrt{3}-1\big)$
$=\sqrt{3}\big[(\sqrt{3}) ^2-1\big]$
$=\sqrt{3}(3-1) =2\sqrt{3}$
$\text{L.H.S}=\text{R.H.S}$
$\Big [\because \tan60^\circ=\sqrt{3},\sin60^\circ=\frac {\sqrt{3}}{2}\text{ and }\cot30^\circ= \sqrt{3}\Big]$
Hence proved.
$=\big(\sqrt{3}\big)^3-2\frac {\sqrt{3}}{2}=3\sqrt{3}-\sqrt{3}=2\sqrt {3}$
$\text{L.H.S}=\big(\sqrt{3}+1\big)\big(3-\cot30^ \circ\big)$
$=\big(\sqrt{3}+1\big)\big(3- \sqrt{3}\big)$
$=\big(\sqrt{3}+1\big)\sqrt {3}\big(\sqrt{3}-1\big)$
$=\sqrt{3}\big[(\sqrt{3}) ^2-1\big]$
$=\sqrt{3}(3-1) =2\sqrt{3}$
$\text{L.H.S}=\text{R.H.S}$
$\Big [\because \tan60^\circ=\sqrt{3},\sin60^\circ=\frac {\sqrt{3}}{2}\text{ and }\cot30^\circ= \sqrt{3}\Big]$
Hence proved.
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