Question

4 cot square 60 degree + sin square 30 degree minus 2 sin squared 45 degree by sin square 60 degree + cos square 45 degree

Answer 1

this is your solution

Answer 2

$Value \:of \: \frac{4cot^{2}60\degree + sin^{2} 30 \degree - 2sin^{2} 45 \degree }{ sin^{2} 60\degree + cos^{2} 45\degree }$

We know the values of the following:

$i ) cot 60 \degree = \frac{1}{\sqrt{3}} \\ii) sin 30 \degree = \frac{1}{2} \\iii ) sin 45\degree = \frac{1}{\sqrt{2}} \\iv ) sin 60 \degree = \frac{\sqrt{3}}{2} \\v) cos 45\degree = \frac{1}{\sqrt{2}}$

$= \frac{4\Big(\frac{1}{\sqrt{3}}\Big)^{2} + \Big(\frac{1}{2}\Big)^{2} - 2\Big(\frac{1}{\sqrt{2}}\Big)^{2} }{ \Big(\frac{\sqrt{3}}{2}\Big)^{2} + \Big(\frac{1}{\sqrt{2}}\Big)^{2} }$

$= \frac{\Big(\frac{4}{3}\Big)+ \Big(\frac{1}{4}\Big)-\Big(\frac{2}{2}\Big)}{\Big(\frac{3}{4}\Big)+ \Big(\frac{1}{2}\Big)}$

$= \frac{ \frac{16+3-12}{12}}{\frac{3+2}{4}}\\= \frac{\frac{7}{12}}{\frac{5}{4}} \\= \frac{7}{12}\times \frac{4}{5}\\= \frac{7}{15}$

Therefore.,

$\red {\frac{4cot^{2}60\degree + sin^{2} 30 \degree - 2sin^{2} 45 \degree }{ sin^{2} 60\degree + cos^{2} 45\degree }}\green {= \frac{7}{15} }$

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