Question 5-3 - 13-cot 30-tan3 60- - 2 sin 60-

RHS=tan^3 60^∘ 2 sin 60^∘ =(√(3))^3 2√(3)/2=3√(3) √(3)=2√(3) LHS=(√(3)+1)(3 cot 30^∘)=(√(3)+1)(3 √(3)). [ ∵ tan 60^∘ = √(3), sin 60^∘=√(3)/2 and cot 30^∘= ...

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Standard X

Mathematics

Trigonometric Table

Question 5√3 ...

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Question 5
$(\sqrt{3} + 1) (3 - c o t 30^{\circ}) = t a n^{3} 60^{\circ} - 2 s i n 60^{\circ}$

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Solution

$R H S = t a n^{3} 60^{\circ} - 2 s i n 60^{\circ} = (\sqrt{3})^{3} - 2 \frac{\sqrt{3}}{2} = 3 \sqrt{3} - \sqrt{3} = 2 \sqrt{3}$

$L H S = (\sqrt{3} + 1) (3 - c o t 30^{\circ}) = (\sqrt{3} + 1) (3 - \sqrt{3})$ .

$[∵ t a n 60^{\circ} = \sqrt{3}, s i n 60^{\circ} = \frac{\sqrt{3}}{2} a n d c o t 30^{\circ} = \sqrt{3}]$

$= (\sqrt{3} + 1) (\sqrt{3} (\sqrt{3} - 1))$
$= \sqrt{3} ({\sqrt{3}}^{2} - 1) = \sqrt{3} (3 - 1) = 2 \sqrt{3}$

$∴ L H S = R H S$

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Q. Question 5

(\sqrt{3} + 1) (3 - c o t 30^{\circ}) = t a n^{3} 60^{\circ} - 2 s i n 60^{\circ}

Q. Prove the following:

(\sqrt{3} + 1) (3 - cot 30^{\circ}) = {tan}^{3} 60^{\circ} - 2 sin 60^{\circ}

Q. Prove that

(\sqrt{3} + 1) (3 - c o t 30^{°}) = \tan^{3} 60^{°} - 2 \sin 60^{°}

Q. Evaluate :

(3 - cot 30^{\circ}) - {tan}^{3} 60^{\circ} + 2 tan 60^{\circ}

\frac{{tan}^{2} 60^{0} + 4 {sin}^{2} 45^{0} + 3 {sec}^{2} 30^{0} + 5 {cos}^{2} 90^{0}}{csc 30^{0} + sec 60^{0} - {cot}^{2} 30^{0}} =

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Trigonometric Table

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Question 5 (√3+1)(3−cot 30∘)=tan3 60∘−2 sin 60∘

RHS=tan3 60∘−2 sin 60∘=(√3)3−2√32=3√3−√3=2√3 LHS=(√3+1)(3−cot 30∘)=(√3+1)(3−√3). [∵tan 60∘=√3, sin 60∘=√32 and cot 30∘=√3] =(√3+1)(√3(√3−1)) =√3(√32−1)=√3(3−1)=2√3 ∴LHS=RHS

Question 5
$(\sqrt{3} + 1) (3 - c o t 30^{\circ}) = t a n^{3} 60^{\circ} - 2 s i n 60^{\circ}$