Question

$\sf\dfrac{4}{cot^230^{\circ} } +\dfrac{1}{sin^2 30^{\circ}} -2 cos^245^{\circ}-sin^2 0^{\circ}$=​

Answer 1

Answer :-

$\bf{\dfrac{13}{ 3} }$

Step-by-step explanation :-

$\sf \dfrac{4}{cot^230^{\circ} } +\dfrac{1}{sin^2 30^{\circ}} -2 cos^245^{\circ}-sin^2 0^{\circ}$

$\displaystyle \sf{=\frac{4}{(\sqrt{3})^2}+\frac{1}{(\frac{1}{2} )^2} -2\times (\frac{1}{\sqrt{2} } )^2-(0)^2 }$

$\displaystyle \sf{=\frac{4}{3}+\frac{1}{1/4}-2\times \frac{1}{2}-0 }$

$\displaystyle \sf{=\frac{4}{3}+4-1 }$

$\displaystyle \sf{=\frac{4}{3}+3 }$

$\displaystyle \sf{=\frac{4+9}{3} }$

$\displaystyle \sf{=\frac{13}{3} }$

Formulas used :-

$\bullet\ \sf{cot30^\circ=\sqrt{3}}$

$\bullet\ \sf{sin30^\circ=\dfrac{1}{2} }$

$\bullet\ \sf{cos45^\circ=\dfrac{1}{\sqrt{2}} }$

$\bullet\ \sf{sin0^\circ=0}$

Answer 2

$\frac{4}{cot^{2}30° } \: + \: \frac{1}{sin^{2}30° } \: - \: 2cos ^{2}45° \: - \: sin^{2}0 \\ \\ = \frac{4}{( \sqrt{3} )^{2} } + \frac{1}{( \frac{1}{2})^{2} } \: - 2 \times ( \frac{1}{ \sqrt{2} } ) ^{2} - 0 \\ \\ = \frac{4}{3} + 4 - 1 \\ \\ = \frac{4}{3} + 3 \\ \\ = \frac{13}{3}$ Hope it's help you.......