Question

If $\sec A=\frac{5}{4}$, verify that $\frac{3\sin A-4{\sin }^{3}A}{4{\cos }^{3}A-3\cos A}=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

Answer 1

Given, $\sec A=\frac{5}{4}=\frac{Hypotenuse}{adjacentSide}$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

$5^2=(oppositeSide{)}^2+4^2$

$(oppositeSide{)}^2=5^2-4^2=25-16$

$(oppositeSide{)}^2=9=3^2$

$\left(oppositeSide\right)=3$

$\sin A=\frac{3}{5},\cos A=\frac{4}{5},\tan A=\frac{3}{4}$

To Prove, $\frac{3\sin A-4{\sin }^{3}A}{4{\cos }^{3}A-3\cos A}=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

$\frac{3\left(\frac{3}{5}\right)-4{\left(\frac{3}{5}\right)}^3}{4{\left(\frac{4}{5}\right)}^3-3\left(\frac{4}{5}\right)}=\frac{3\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^3}{1-3{\left(\frac{3}{4}\right)}^2}$

$\frac{\left(\frac{9}{5}\right)-\left(\frac{108}{125}\right)}{\left(\frac{256}{125}\right)-\left(\frac{12}{5}\right)}=\frac{\left(\frac{9}{4}\right)-\left(\frac{27}{64}\right)}{1-\left(\frac{27}{16}\right)}$

$\frac{\left(\frac{225-108}{125}\right)}{\left(\frac{256-300}{125}\right)}=\frac{\left(\frac{144-27}{64}\right)}{\left(\frac{16-27}{16}\right)}$

$\frac{\left(\frac{117}{125}\right)}{\left(\frac{-44}{125}\right)}=\frac{\left(\frac{117}{64}\right)}{\left(\frac{-11}{16}\right)}$

$\frac{\left(\frac{117}{125}\right)}{\left(\frac{-44}{125}\right)}=\frac{\left(\frac{117}{64}\right)}{\left(\frac{-11}{16}\right)}$

$\frac{117}{-44}=\frac{117}{11(-4)}$

$\frac{117}{-44}=\frac{117}{-44}$

$\therefore \frac{3\sin A-4{\sin }^{3}A}{4{\cos }^{3}A-3\cos A}=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

Hence Verified