सिद्ध कीजिए $(\sqrt{3} + 1) (3 - \cot 30^\circ) = \tan^3 60^\circ - 2 \sin 60^\circ$
Exercise-8.3-5
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हमें यह साबित करना होगा: $(\sqrt{3} + 1)(3 - \cot 30^\circ) = \tan^3 60^\circ - 2 \tan 60^\circ$
यहाँ, $\text{LHS} = (\sqrt{3} + 1)(3 - \cot 30^\circ)$
$= (\sqrt{3}+1)(3-\sqrt{3})$
$= \sqrt{3}(3-\sqrt{3})+1(3-\sqrt{3})$
$= 3 \sqrt{3}-3+3-\sqrt{3}$
$= 2 \sqrt{3}$
$\text{RHS} = \tan^3 60^\circ - 2 \tan 60^\circ$
$= (\sqrt{3})^{3}-2 \times \frac{\sqrt{3}}{2}$
$= 3 \sqrt{3}-\sqrt{3}$
$= 2 \sqrt{3}$
$\text{LHS = RHS}$
सिद्ध किया।
यहाँ, $\text{LHS} = (\sqrt{3} + 1)(3 - \cot 30^\circ)$
$= (\sqrt{3}+1)(3-\sqrt{3})$
$= \sqrt{3}(3-\sqrt{3})+1(3-\sqrt{3})$
$= 3 \sqrt{3}-3+3-\sqrt{3}$
$= 2 \sqrt{3}$
$\text{RHS} = \tan^3 60^\circ - 2 \tan 60^\circ$
$= (\sqrt{3})^{3}-2 \times \frac{\sqrt{3}}{2}$
$= 3 \sqrt{3}-\sqrt{3}$
$= 2 \sqrt{3}$
$\text{LHS = RHS}$
सिद्ध किया।
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