Question

\frac{5sin^{2}(30)^{o}+cos^{2}(45)^{o}-4tan^{2}(30)^{o})}
{2sin(30^{o})cos(30^{o})+tan(45^{o})}​

Answer 1

Correct Question »

$\displaystyle\bf{\dfrac{5\sin^230°+\cos^245°-4\tan^230°}{2\sin30°.\cos30°+\tan45°}}$

Analysis→

Here we're given an equation $\displaystyle\rm{\dfrac{5\sin^230°+\cos^245°-4\tan^230°}{2\sin30°.\cos30°+\tan45°}}$ And we've to find the value of the equation. And we know that :

$\dashrightarrow\rm{\sin30°=\dfrac{1}{2}}$

$\dashrightarrow\rm{\cos30°=\dfrac{\sqrt{3}}{2}}$

$\dashrightarrow\rm{\cos45°=\dfrac{1}{\sqrt{2}}}$

$\dashrightarrow\rm{\tan30°=\dfrac{1}{\sqrt{3}}}$

$\dashrightarrow\rm{\tan45°=1}$

Given→

$\displaystyle\rm{Equation=\dfrac{5\sin^230°+\cos^245°-4\tan^230°}{2\sin30°.\cos30°+\tan45°}}$

To Find→

The value of the given equation.

Answer→

$\displaystyle\rm{\dfrac{5\sin^230°+\cos^245°-4\tan^230°}{2\sin30°.\cos30°+\tan45°}}$

$\displaystyle\implies\rm{\dfrac{5\Big(\dfrac{1}{2}\Big)^2+\Big(\dfrac{1}{\sqrt{2}}\Big)^2-4\Big(\dfrac{1}{\sqrt{3}}\Big)^2}{2\Big(\dfrac{1}{2}\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{5\Big(\dfrac{1}{4}\Big)+\Big(\dfrac{1}{2}\Big)-4\Big(\dfrac{1}{3}\Big)}{\cancel{2}\Big(\dfrac{1}{\cancel{2}}\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{4}\Big)+\Big(\dfrac{1}{2}\Big)-\Big(\dfrac{4}{3}\Big)}{\Big(1\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\rm{\dfrac{\Big(\dfrac{15}{12}\Big)+\Big(\dfrac{6}{12}\Big)-\Big(\dfrac{16}{12}\Big)\qquad\big[LCM(2,3,4=15)\big]}{\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{15+6-16}{12}\Big)}{\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{12}\Big)}{\Big(\dfrac{\sqrt{3}+2}{2}\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{\cancel{12}}\Big)}{\Big(\dfrac{\sqrt{3}+2}{\cancel{2}}\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{6}\Big)}{\Big(\sqrt{3}+2\Big)}}$

$\displaystyle\implies\rm{\dfrac{5}{6\Big(\sqrt{3}+2\Big)}}$

$\displaystyle\implies\rm{\dfrac{5}{6\sqrt{3}+12}}$

${\boxed{\boxed{\implies{\bf{\dfrac{5}{6\sqrt{3}+12}\checkmark}}}}}$

☞ Hence the value of the given equation is $\displaystyle\rm{\dfrac{5}{6\sqrt{3}+12}}$ which is the required answer.

HOPE IT HELPS.

Answer 2

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●▬▬▬▬๑۩۩๑▬▬▬▬▬●

$\frac{5sin^{2}(30)^{o}+cos^{2}(45)^{o}-4tan^{2}(30)^{o})}</p><p>{2sin(30^{o})cos(30^{o})+tan(45^{o})}$

●▬▬▬▬๑۩۩๑▬▬▬▬▬●

$\huge\boxed{\fcolorbox{black}{red}{ƛnsweR}}$

Explanation:

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$\displaystyle\implies\rm{\dfrac{5\Big(\dfrac{1}{2}\Big)^2+\Big(\dfrac{1}{\sqrt{2}}\Big)^2-4\Big(\dfrac{1}{\sqrt{3}}\Big)^2}{2\Big(\dfrac{1}{2}\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{5\Big(\dfrac{1}{4}\Big)+\Big(\dfrac{1}{2}\Big)-4\Big(\dfrac{1}{3}\Big)}{\cancel{2}\Big(\dfrac{1}{\cancel{2}}\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{4}\Big)+\Big(\dfrac{1}{2}\Big)-\Big(\dfrac{4}{3}\Big)}{\Big(1\Big).\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\rm{\dfrac{\Big(\dfrac{15}{12}\Big)+\Big(\dfrac{6}{12}\Big)-\Big(\dfrac{16}{12}\Big)\qquad\big[LCM(2,3,4=15)\big]}{\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{15+6-16}{12}\Big)}{\Big(\dfrac{\sqrt{3}}{2}\Big)+\Big(1\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{12}\Big)}{\Big(\dfrac{\sqrt{3}+2}{2}\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{\cancel{12}}\Big)}{\Big(\dfrac{\sqrt{3}+2}{\cancel{2}}\Big)}}$

$\displaystyle\implies\rm{\dfrac{\Big(\dfrac{5}{6}\Big)}{\Big(\sqrt{3}+2\Big)}}$

$\displaystyle\implies\rm{\dfrac{5}{6\Big(\sqrt{3}+2\Big)}}$

$\displaystyle\implies\rm{\dfrac{5}{6\sqrt{3}+12}}$

${\boxed{\boxed{\implies{\bf{\dfrac{5}{6\sqrt{3}+12}\checkmark}}}}}$

☞ Hence the value of the given equation is $\displaystyle\rm{\dfrac{5}{6\sqrt{3}+12}}$which is the required answer.

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$\rm\underline\purple{Hope \: its \: help \: you \:☺}$

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