Question

cot square 30 degree subtract 2 cos square 60 degree minus 3 upon 4 sec square 45 degree minus 4 sec square 30 degree​

Answer 1

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Answer 2

$\dfrac{\cot^2 30-2\cos^2 60-3}{4\sec^245-4\sin^2 30}=-\dfrac{1}{14}$

Step-by-step explanation:

Given : Expression $\dfrac{\cot^2 30-2\cos^2 60-3}{4\sec^245-4\sin^2 30}$

To find : Solve the expression ?

Solution :

Expression $\dfrac{\cot^2 30-2\cos^2 60-3}{4\sec^245-4\sin^2 30}$

Using trigonometric values,

$\cot 30=\sqrt{3}$

$\cos 60=\frac{1}{2}$

$\sec 45=\sqrt{2}$

$\sin 30=\frac{1}{2}$

Substitute the values,

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

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

$=\dfrac{-\frac{1}{2}}{8-1}$

$=\dfrac{-\frac{1}{2}}{7}$

$=-\dfrac{1}{14}$

Therefore, the value of given expression is$\dfrac{\cot^2 30-2\cos^2 60-3}{4\sec^245-4\sin^2 30}=-\dfrac{1}{14}$.

#Learn more

Cot square 30 degree subtract 2 cos square 60 degree minus 3 upon 4 sec square 45 degree minus 4 sec square 30 degree​.

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