Question

4 cot 3 60+9sin2 60-6cosec2 60-9/4tan2 60=4

Answer 1

Answer:

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Answer 2

Answers:

$\huge{\boxed{\sf{671.12}}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\frac{4 \cot^{3} (60) + 9 \sin^{2} (60) - 6 \cosec^{2} (60) - 9}{4 \tan^{2} (60) } = 4 \\ = )\frac{4 \times ( \frac{1}{ \sqrt{3} })^{2} + 9 \times ( \frac{ \sqrt{3} }{2})^{2} - 6 \times( { \frac{2}{ \sqrt{3} } })^{2} - 9 }{4 \times { \sqrt{3} }^{2} } = 4 \\ = ) \frac{4 \times \frac{1}{3} \times \frac{1}{ \sqrt{3}} +{ 9 \times \frac{3}{4} } - 6 \times \frac{4}{3} - 9 }{4 \times 3} = 4 \\ = ) \frac{ \frac{4}{3 \sqrt{3} } + \frac{27}{4} - \frac{24}{3} - \frac{9}{1} }{12 } = 4 \\ = ) \frac{4}{3 \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } + \frac{27}{4} - \frac{24}{3} - \frac{9}{1} = 48 \\ = ) \frac{12 \sqrt{3} }{9} + \frac{27}{4} - \frac{24}{3} - \frac{9}{1} = 48 \\ = ) \frac{4 \sqrt{3} }{3} + \frac{27}{4} - \frac{24}{3} - \frac{9}{1} = 48 \\ = ) \frac{16 \sqrt{3} +8 1 - 96 - 108}{12} = 48 \\ = )16 \sqrt{3} + 81 - 204 = 576 \\ = )0 = 576 + 204 - 81 - 16 \sqrt{3} \\ = )780 - 81 - 16 \times 1.73 \\ = )699 - 27.88 \\ = )671.12$

$\huge{\boxed{\sf{671.12}}}$