Question #758df

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Question #758df

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Answer 1 · 2017-04-17T15:19:28

$tan (\frac{11 π}{3}) - 2 sin (\frac{4 π}{6}) - (\frac{3}{4}) cos e c^{2} (\frac{π}{4}) + 4 {cos}^{2} (\frac{17 π}{6})$ $tan((11pi)/3)-2sin((4pi)/6)-(3/4)cosec^2(pi/4)+4cos^2((17pi)/6)$

$= tan (4 π - \frac{π}{3}) - 2 sin (π - \frac{π}{3}) - (\frac{3}{4}) cos e c^{2} (\frac{π}{4}) + 4 {cos}^{2} (3 π - \frac{π}{6})$ $=tan(4pi-pi/3)-2sin(pi-pi/3)-(3/4)cosec^2(pi/4)+4cos^2(3pi-pi/6)$

$= - tan (\frac{π}{3}) - 2 sin (\frac{π}{3}) - (\frac{3}{4}) cos e c^{2} (\frac{π}{4}) + 4 {cos}^{2} (\frac{π}{6})$ $=-tan(pi/3)-2sin(pi/3)-(3/4)cosec^2(pi/4)+4cos^2(pi/6)$

$= - \sqrt{3} - 2 \times \frac{\sqrt{3}}{2} - (\frac{3}{4}) \times {(\sqrt{2})}^{2} + 4 {(\frac{\sqrt{3}}{2})}^{2}$ $=-sqrt3-2xxsqrt3/2-(3/4)xx(sqrt2)^2+4(sqrt3/2)^2$

$= - \sqrt{3} - \sqrt{3} - (\frac{3}{4}) \times 2 + 4 \times \frac{3}{4}$ $=-sqrt3-sqrt3-(3/4)xx2+4xx3/4$

$= - \sqrt{3} - \sqrt{3} - \frac{3}{2} + 3$ $=-sqrt3-sqrt3-3/2+3$

$= \frac{3}{2} - 2 \sqrt{3}$ $=3/2-2sqrt3$

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