Question

evaluate the following 5sin 30 degree -cos^2 45 degree - 4 tan 30 degree /2 sin 30 degree cos 30 degree - tan 45 degree
please it's very important ​

Answer 1

★ Given . . .

$\:$

• $\sf \dfrac{5\:sin30^{\circ} - cos^{2}45^{\circ} - 4\:tan30^{\circ}}{2\:sin30^{\circ}.cos30^{\circ} - tan45^{\circ}}$

$\:$

★ What to do . . .

$\:$

⚝ Evaluate :

$\:$

• $\sf \dfrac{5\:sin30^{\circ} - cos^{2}45^{\circ} - 4\:tan30^{\circ}}{2\:sin30^{\circ}.cos30^{\circ} - tan45^{\circ}}$

$\:$

★ Required Solution . . .

$\\ :\implies\:\sf \dfrac{ \Bigg \{5\:\times\:\dfrac{1}{2}\Bigg\} - \Bigg\{\dfrac{1}{\sqrt{2}}\Bigg\}^2 - \Bigg\{4\:\times\:\dfrac{1}{\sqrt{3}}\Bigg\}}{2\:\times\:\dfrac{1}{2}\:\times\:\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{5}{2} - \dfrac{1}{2} - \dfrac{4}{\sqrt{3}}}{\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{5 - 1}{2} - \dfrac{4}{\sqrt{3}}}{\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{4}{2} - \dfrac{4}{\sqrt{3}}}{\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{4}{2} - \dfrac{4}{\sqrt{3}}}{\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{4}{2} - \dfrac{4}{\sqrt{3}}\:\times\:\dfrac{\sqrt{3}}{\sqrt{3}}}{\dfrac{2}{2}\:\times\:\dfrac{\sqrt{3}}{2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{4}{2} - \dfrac{4\sqrt{3}}{(\sqrt{3})^2}}{\dfrac{2\sqrt{3}}{(2)^2} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{4}{2} - \dfrac{4\sqrt{3}}{3}}{\dfrac{2\sqrt{3}}{4} - 1}$

$\\ :\implies\:\sf \dfrac{\dfrac{12 - 8\sqrt{3}}{6}}{\dfrac{2\sqrt{3} - 4}{4}}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{6}\:\div\:\dfrac{2\sqrt{3} - 4}{4}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{\cancel{6}}\:\times\:\dfrac{\cancel{4}}{2\sqrt{3} - 4}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{3}\:\times\:\dfrac{2}{2\sqrt{3} - 4}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{3}\:\times\:\dfrac{2}{2\sqrt{3} - 4}\:\times\:\dfrac{2\sqrt{3} + 4}{2\sqrt{3} + 4}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{3}\:\times\:\dfrac{2(2\sqrt{3} + 4)}{(2\sqrt{3})^2 - (4)^2}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{3}\:\times\:\dfrac{8 + 4\sqrt{3}}{12 - 16}$

$\\ :\implies\:\sf \dfrac{12 - 8\sqrt{3}}{3}\:\times\:\dfrac{8 + 4\sqrt{3}}{-4}$

$\\ :\implies\:\sf \dfrac{(12 - 8\sqrt{3})\:\times\:(8 + 4\sqrt{3})}{3\:\times\:(-4)}$

$\\ :\implies\:\sf \dfrac{(12 - 8\sqrt{3})\:\times\:(8 + 4\sqrt{3})}{-12}$

$\\ :\implies\:\sf -\dfrac{(12 - 8\sqrt{3})\:\times\:(8 + 4\sqrt{3})}{12}$

$\\ :\implies\:\sf -\dfrac{4(3 - 2\sqrt{3})\:\times\:(2 + \sqrt{3})}{3}$

$\\ :\implies\:\sf -\dfrac{-4\sqrt{3}}{3}$

$\\ :\implies\:\sf -\Bigg[-\dfrac{4\sqrt{3}}{3}\Bigg]$

$\\ :\implies\:\sf \red{\dfrac{4\sqrt{3}}{3}}$

Hence,

$\boxed{\bf{\dfrac{5\:sin30^{\circ} - cos^{2}45^{\circ} - 4\:tan30^{\circ}}{2\:sin30^{\circ}.cos30^{\circ} - tan45^{\circ}} = \dfrac{4\sqrt{3}}{3}}}$

★ Trigonometry table :

$\large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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Answer 2

Answer:

Answer is 4√3/3

Step-by-step explanation:

To evaluate :

$\star \sf \: \dfrac{5 \: sin \: 30\degree - cos^{2} \: 45\degree - 4 \: tan \: 30\degree}{2 \: sin \: 30\degree \:cos \: 30\degree - tan \: 45\degree}$

Solution :

We know values :

Sin 30° = 1/2

Cos 45° = 1/√2

Tan 30° = 1/√3

Cos 30° √3/2

Tan 45° = 1

Put the values :

$\longrightarrow \sf \dfrac{5 \times \bigg( \dfrac{1}{2} \bigg) - \bigg(\dfrac{1}{\sqrt{2}}\bigg) ^{2} - 4 \times \bigg( \dfrac{1}{\sqrt{3}}\bigg)}{2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} - 1}$

$\longrightarrow \sf \dfrac{\dfrac{5}{2} - \dfrac{1}{2} - \dfrac{4}{\sqrt{3}}}{\dfrac{2}{2} \times \dfrac{\sqrt{3}}{2} - 1}$

$\longrightarrow \sf \dfrac{\dfrac{5-1}{2} - \dfrac{4}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}}{\dfrac{2\sqrt{3}}{4} - 1}$

$\longrightarrow \sf \dfrac{\dfrac{4}{2} - \dfrac{4 \sqrt{3}}{3}}{\dfrac{2 \: \sqrt{3} - 4}{4}}$

$\longrightarrow \sf \dfrac{\dfrac{12 - 8 \: \sqrt{3}}{\cancel{6}}}{\dfrac{2 \sqrt{3} - 4}{\cancel{4}}}$

$\longrightarrow \sf \dfrac{\dfrac{12 - 8 \: \sqrt{3}}{3}}{\dfrac{2 \sqrt{3} - 4}{2}}$

$\longrightarrow \sf \dfrac{12 - 8\sqrt{3}}{3}$ ÷ $\sf \dfrac{2\sqrt{3} - 4}{2}$

$\longrightarrow \sf \dfrac{12 - 8 \sqrt{3}}{3} \times \dfrac{2}{2\sqrt{3} - 4}$

$\longrightarrow \sf \dfrac{12 - 8\sqrt{3}}{3} \times \bigg(\dfrac{2}{2 \sqrt{3} - 4} \times \dfrac{2 \sqrt{3} + 4}{2 \sqrt{3} + 4} \bigg)$

$\longrightarrow \sf \dfrac{12 - 8\sqrt{3}}{3} \times \dfrac{8 + 4\sqrt{3}}{(2\sqrt{3})^{2} - (4)^{2}}$

$\longrightarrow \sf \dfrac{12 - 8\sqrt{3}}{3} \times \dfrac{8 + 4 \sqrt{3}}{-4}$

$\longrightarrow \sf \dfrac{(12 - 8\sqrt{3}) \times (8 + 4 \sqrt{3})}{-12}$

$\longrightarrow \sf \dfrac{12(8 + 4\sqrt{3}) - 8\sqrt{3}(8 + 4\sqrt{3})}{-12}$

$\longrightarrow \sf \dfrac{\cancel{96} + 48\sqrt{3} - 64\sqrt{3} \cancel{ - 96}}{-12}$

$\longrightarrow \sf \dfrac{-16\sqrt{3}}{-12}$

$\longrightarrow$ 4√3/3

Therefore,

$\star \sf \: \dfrac{5 \: sin \: 30\degree - cos^{2} \: 45\degree - 4 \: tan \: 30\degree}{2 \: sin \: 30\degree \:cos \: 30\degree - tan \: 45\degree} = \dfrac{4\sqrt{3}}{3}$