Question

4/cot^2 30°+ 1/sin^2 60°- cos^2 45°

Answer 1

Answer:

$\frac{4}{cot^{2}30\degree}+\frac{1}{sin^{2}60\degree}-cos^{2}45\degree=\frac{13}{6}$

Step-by-step explanation:

$\frac{4}{cot^{2}30\degree}+\frac{1}{sin^{2}60\degree}-cos^{2}45\degree$

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

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

$=\frac{8+8-3}{6}$

$=\frac{13}{6}$

Therefore,

$\frac{4}{cot^{2}30\degree}+\frac{1}{sin^{2}60\degree}-cos^{2}45\degree=\frac{13}{6}$

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Answer 2

Given:

$\frac{4}{cot^2 30^{\circ}}+\frac{1}{sin^260^{\circ}}-cos^2 45^{\circ}$

To find:

Value of $\frac{4}{cot^2 30}+\frac{1}{sin^260}-cos^2 45$

Solution:

$\frac{4}{cot^2 30}+\frac{1}{sin^260}-cos^2 45$

We know that

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

$sin60^{\circ}=\frac{\sqrt{3}}{2}$

$cos45^{\circ}=\frac{1}{\sqrt{2}}$

Using the values then we get

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

$\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$

$\frac{4+4}{3}-\frac{1}{2}$

$\frac{8}{3}-\frac{1}{2}$

$\frac{16-3}{6}$

$\frac{13}{6}$

$\frac{4}{cot^2 30}+\frac{1}{sin^260}-cos^2 45=\frac{13}{6}$