Evaluate- 2sin230o 30o-3cos260o 230o-

2sin 30^∘30^∘ 3cos^260^∘^230^∘ =2(12)^2(√(3)) 3(12)^2(2√(3))^2 =√(3)2 1 ...

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Byju's Answer

Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

Evaluate: 2...

Question

Evaluate: $2 {sin}^{2} 30^{o} cot 30^{o} - 3 {cos}^{2} 60^{o} {sec}^{2} 30^{o}$ .

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Solution

$2 sin 30^{\circ} cot 30^{\circ} - 3 {cos}^{2} 60^{\circ} {sec}^{2} 30^{\circ}$
$= 2 (\frac{1}{2})^{2} (\sqrt{3}) - 3 (\frac{1}{2})^{2} (\frac{2}{\sqrt{3}})^{2}$
$= \frac{\sqrt{3}}{2} - 1$

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Evaluate: 2sin230o cot30o−3cos260o sec230o.

2sin30∘cot30∘−3cos260∘sec230∘=2(12)2(√3)−3(12)2(2√3)2=√32−1

Evaluate: $2 {sin}^{2} 30^{o} cot 30^{o} - 3 {cos}^{2} 60^{o} {sec}^{2} 30^{o}$ .

$2 sin 30^{\circ} cot 30^{\circ} - 3 {cos}^{2} 60^{\circ} {sec}^{2} 30^{\circ}$
$= 2 (\frac{1}{2})^{2} (\sqrt{3}) - 3 (\frac{1}{2})^{2} (\frac{2}{\sqrt{3}})^{2}$
$= \frac{\sqrt{3}}{2} - 1$