Question

evaluate: cos²60°+4sec²30°- tan²45°/sin²30°+cos²30°​

Answer 1

To find:

$\sf{The \ value \ of \ \dfrac{cos^{2}60^\circ+4sec^{2}30^\circ-tan^{2}45^\circ}{sin^{2}30^\circ+cos^{2}30^\circ}}$

Solution:

$\sf{Note:}$

$\sf{cos60^\circ=\dfrac{1}{2},}$

$\sf{sec30^\circ=\dfrac{2}{\sqrt3},}$

$\sf{tan45^\circ=1,}$

$\sf{sin30^\circ=\dfrac{1}{2},}$

$\sf{cos30^\circ=\dfrac{\sqrt3}{2}}$

$\sf{\leadsto{\dfrac{cos^{2}60^\circ+4sec^{2}30^\circ-tan^{2}\circ}{sin^{2}30^\circ+cos^{2}30^\circ}}}$

$\sf{Substituting \ the \ values}$

$\sf{\leadsto{\dfrac{(\frac{1}{2})^{2}+4\times(\frac{2}{\sqrt3})^{2}-(1)^{2}}{(\frac{1}{2})^{2}+(\frac{\sqrt3}{2})^{2}}}}$

$\sf{\leadsto{\dfrac{\frac{1}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}}}$

$\sf{\leadsto{\dfrac{\frac{3+64-12}{12}}{\frac{4}{4}}}}$

$\sf{\leadsto{\dfrac{55}{12}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \dfrac{cos^{2}60^\circ+4sec^{2}30^\circ-tan^{2}45^\circ}{sin^{2}30^\circ+cos^{2}30^\circ}}}}$

$\sf\purple{\tt{is \ \dfrac{55}{12}.}}$