Question

If sec $A=\frac{5}{4}$, verify that $\frac{3sinA-4si{n}^{3}A}{4co{s}^{3}A-3cosA}=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$

Answer 1

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

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

By pythagoras theorem,

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

⇒Prependicluar = $\sqrt{25-16}$

⇒Prependicluar = 3

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

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

Now, we will prove that

$\frac{3sinA-4si{n}^{3}A}{4co{s}^{3}A-3cosA}=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$

LHS = $\frac{3sinA-4si{n}^{3}A}{4co{s}^{3}A-3cosA}$

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

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

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

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

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

RHS = $\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$

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

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

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

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

$\therefore LHS=RHS$ Hence proved.