Question

[5sin²30°+cos²45°+4tan²60°] /(2sin30°cos60°+tan45°)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\dfrac{5\sin^2 30^\circ+\cos^2 45^\circ+4\tan^2 60^\circ}{2\sin 30^\circ\cos 60^\circ+\tan 45^\circ}=\dfrac{55}{6}$

Step-by-step explanation:

Given : Expression $\dfrac{5\sin^2 30^\circ+\cos^2 45^\circ+4\tan^2 60^\circ}{2\sin 30^\circ\cos 60^\circ+\tan 45^\circ}$

To find : Simplify the expression ?

Solution :

$\dfrac{5\sin^2 30^\circ+\cos^2 45^\circ+4\tan^2 60^\circ}{2\sin 30^\circ\cos 60^\circ+\tan 45^\circ}$

Using trigonometric values,

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

$\cos 45^\circ=\frac{1}{\sqrt2}$

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

$\tan 45^\circ=1$

$\tan 60^\circ=\sqrt{3}$

Substitute the values,

$=\dfrac{5(\frac{1}{2})^2+(\frac{1}{\sqrt2})^2+4(\sqrt{3})^2}{2(\frac{1}{2})(\frac{1}{2})+1}$

$=\dfrac{\frac{5}{4}+\frac{1}{2}+12}{\frac{1}{2}+1}$

$=\dfrac{\frac{5+2+48}{4}}{\frac{1+2}{2}}$

$=\dfrac{\frac{55}{4}}{\frac{3}{2}}$

$=\dfrac{55}{6}$

Therefore, $\dfrac{5\sin^2 30^\circ+\cos^2 45^\circ+4\tan^2 60^\circ}{2\sin 30^\circ\cos 60^\circ+\tan 45^\circ}=\dfrac{55}{6}$

#Learn more

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