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

We have,sec $A=\frac{5}{4}$

$\Rightarrow cosA=\frac{4}{5}$ [$\frac{Base}{Hypotenuse}$]

By pythagoras theorem,

$(Perpendicular{)}^2=(Hypotenuse{)}^2-(Base{)}^2$

⇒Perpendicular = $\sqrt{25-16}$

⇒Perpendicular = 3

Then, sin A = $\frac{Perpendicular}{Hypotenuse}=\frac{3}{5}$

tan A = $\frac{Perpendicular}{Base}=\frac{3}{4}$

Now, we will prove that

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

LHS = $\frac{3\sin A-4{\sin }^{3}A}{4{\cos }^{3}A-3\cos A}$

$=\frac{3\times \frac{3}{5}-4(\frac{3}{5}{)}^3}{4(\frac{4}{5}{)}^3-3\times \frac{4}{5}}$

$\frac{\frac{9}{5}-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}$

$=\frac{\frac{225-108}{125}}{\frac{256-300}{125}}$

$=\frac{117}{125}\times \frac{125}{-44}$

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

RHS = $\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

$=\frac{3\times \frac{3}{4}-(\frac{3}{4}{)}^3}{1-3(\frac{3}{4}{)}^2}$

$=\frac{\frac{9}{4}-\frac{27}{64}}{1-\frac{27}{16}}$

$=\frac{\frac{117}{64}}{\frac{-11}{16}}$

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

$\therefore LHS=RHS$