Question

find the value of the above ​

Answer 1

Step-by-step explanation:

3+4+2+0/4+6-7

9/3

=3

3 is the answer

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Answer 2

$\bf{Given}$

$\sf\to \dfrac{Cot^2 30^{\circ}+8Sin^2 45^{\circ}+\dfrac{3}{2}Sec^2 30^{\circ}+2Cos^2 90^{\circ} }{2Sec30^{\circ} + 3Cosec30^{\circ}-\dfrac{7}{3} Tan^2 60}$

$\bf{We\:know\:that$

$\sf\to Cot30^{\circ}=\sqrt{3}$

$\sf\to Sin45^{\circ}=\dfrac{1}{\sqrt{2} }$

$\sf\to Sec30^{\circ}=\dfrac{2}{\sqrt{3} }$

$\sf\to Cos90^{\circ}=0$

$\sf\to Cosec30^{\circ}=2$

$\sf\to Tan60^{\circ}=\sqrt{3}$

$\bf Now\:\:put\:\:the\:\:value\:\:on\:\:given\:\:equation$

$\sf\to \dfrac{(\sqrt{3} )^2+8\times\bigg(\dfrac{1}{\sqrt{2} }\bigg)^2+\dfrac{3}{2} \times\bigg(\dfrac{2}{\sqrt{3} }\bigg)^2+2\times0 }{2\times\dfrac{2}{\sqrt{3} } +3\times2-\dfrac{7}{3}(\sqrt{3} )^2 }$

$\sf\to\dfrac{3+\dfrac{8}{2}+\dfrac{3}{2} \times \dfrac{4}{3} }{\dfrac{4}{\sqrt{3} }\times\dfrac{\sqrt{3} }{\sqrt{3} } +6-\dfrac{7}{3} \times 3 }$

$\sf\to\dfrac{3+4+2}{\dfrac{4\sqrt{3} }{3}+6-7 }$

$\sf\to\dfrac{9}{\dfrac{4\sqrt{3} }{3} -1}$

$\sf\to\dfrac{9}{\dfrac{4\sqrt{3}-3}{3} }$

$\sf\to\dfrac{27}{4\sqrt{3} - 3}$

$\bf Answer$

$\sf\to\dfrac{27}{4\sqrt{3} - 3}$