Question

4 into sin square 60 degree minus cos square 45 degree by tansquare 30 degree + cos square 90 degree​

Answer 1

Answer:

your solution is as given in the attachment below

Step-by-step explanation:

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Answer 2

Answer:

$\frac{4sin^{2}60-cos^{2}45}{tan^{2}30+cos^{2}90}=\frac{15}{2}$

Step-by-step explanation:

$\frac{4sin^{2}60-cos^{2}45}{tan^{2}30+cos^{2}90}$

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

$=\frac{4\times \frac{3}{4}-\frac{1}{2}}{\frac{1}{3}}$

$=\frac{3-\frac{1}{2}}{\frac{1}{3}}$

$=\frac{6-1}{2}\times \frac{3}{1}\\=\frac{5}{2}\times 3\\=\frac{15}{2}\\=7.5$

Therefore,

$\frac{4sin^{2}60-cos^{2}45}{tan^{2}30+cos^{2}90}=\frac{15}{2}$

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