Math, asked by LegendaryPranav, 9 months ago

4/3 cot²30° + 3 sin²30° - 2 cosec²60° - 3/4 tan²30°

karo Yaar koi to solve....

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Answered by BrainlyNisha001

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$\cot 30 = \sqrt{3} \\ \sin \: 60 = \frac{ \sqrt{3} }{2} \\ \cos \sec \: 60 = \frac{2}{ \sqrt{3} } \\ \tan \: 30 = \frac{1}{ \sqrt{3} }$

$\frac{4}{3} \cot^{2} 30 + 2 \sin ^{2} 60 - 2 \cos\sec^{2} 60 - \frac{3}{4} \tan^{2} 30 \\ = \frac{4}{3} \times ( { \sqrt{3} })^{2} + 2 \times ( \frac{3}{2})^{2} - 2 \times \frac{2}{ \sqrt{3} })^{2} - \frac{3}{4} \times ( \frac{1}{ \sqrt{3} } ) ^{2} \\ = \frac{4}{3} \times 3 + 2 \times \frac{3}{4} - 2 \times \frac{4}{3} - \frac{3}{4} \times \frac{1}{3} \\ = 4 + \frac{3}{2} - \frac{8}{3} - \frac{1}{4} \\ = 4 + 18 - 32 - \frac{3}{12} \\ = 22 - \frac{35}{12} \\ = \frac{ - 13}{12}$

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Answered by pakhi31

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4/3 cot²30° + 3 sin²30° - 2 cosec²60° - 3/4 tan²30°karo Yaar koi to solve....​

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4/3 cot²30° + 3 sin²30° - 2 cosec²60° - 3/4 tan²30°

karo Yaar koi to solve....