Question

Evaluate 2cos 2 60+3sec 2 30-2 tan 2 45 / sin 2 30 +cos 2 45

Answer 1

Solution is attached below

Answer 2

$\dfrac{2\cos^2 60+3\sec^2 30-2\tan^22 45}{\sin^2 30+\cos^2 45}=\dfrac{10}{6}$

Step-by-step explanation:

We have,

$\dfrac{2\cos^2 60+3\sec^2 30-2\tan^22 45}{\sin^2 30+\cos^2 45}$

To find, $\dfrac{2\cos^2 60+3\sec^2 30-2\tan^22 45}{\sin^2 30+\cos^2 45}=?$

$\dfrac{2\cos^2 60+3\sec^2 30-2\tan^22 45}{\sin^2 30+\cos^2 45}$

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

[ ∵ $\cos 60=\dfrac{1}{2}, \sec 30=\dfrac{2}{\sqrt{3}},\tan 45=1$]

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

$=\dfrac{\dfrac{1}{2}+4-2 }{\dfrac{1+2}{4}}$

$=\dfrac{\dfrac{1}{2}+2 }{\dfrac{3}{4}}$

$=\dfrac{\dfrac{5}{2}}{\dfrac{3}{4}}$

$=\dfrac{5}{2} \times \dfrac{4}{3} =\dfrac{10}{6}$

Hence, $\dfrac{2\cos^2 60+3\sec^2 30-2\tan^22 45}{\sin^2 30+\cos^2 45}=\dfrac{10}{6}$