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8. Introduction to Trigonometry
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Prove that,
$(\sqrt{3}+ 1) \left(3-\cot 30^{\circ}\right)=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$
Option A
Option B
Option C
Option D
Solution
R.H.S. $=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}=(\sqrt{3})^{3}-2 \frac{\sqrt{3}}{2}=3 \sqrt{3}-\sqrt{3}=2 \sqrt{3}$
L.H.S. $=(\sqrt{3}+1)\left(3-\cot 30^{\circ}\right)=(\sqrt{3}+1) (3-\sqrt{3})$
$\left[\because \tan 60^{\circ}=\sqrt{3} \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right.$ and $\left.=(\sqrt{3}+1) \sqrt{3}(\sqrt{3}-1) \cot 30^{\circ}=\sqrt{3}\right]$
$=\sqrt{3}(\sqrt{3})^{2}-1=\sqrt{3}(3-1)=2 \sqrt{3}$
L.H.S. $=$ R.H.S.
Standard 10
Mathematics
