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

Given:

$sec\ A = \frac{5}{4}$.

To do:

We have to verify that $\frac{3 \sin A-4 \sin ^{3} A}{4 \cos ^{3} A-3 \cos A}=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$.

Solution:

Let, in a triangle $ABC$ right-angled at $B$, $sec\ A=\frac{5}{4}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (5)^2=(4)^2+BC^2$

$\Rightarrow BC^2=25-16$

$\Rightarrow BC=\sqrt{9}=3$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{3}{5}$

$cos\ \theta=\frac{AB}{AC}=\frac{4}{5}$

$tan\ \theta=\frac{BC}{AB}=\frac{3}{4}$

This implies,

Let us consider LHS,

$\frac{3 \sin A-4 \sin ^{3} A}{4 \cos ^{3} A-3 \cos A}=\frac{3\left(\frac{3}{5}\right) -4\left(\frac{3}{5}\right)^{3}}{4\left(\frac{4}{5}\right)^{3} -3\left(\frac{4}{5}\right)}$

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

$=\frac{\frac{9}{5} -\left(\frac{108}{125}\right)}{\left(\frac{256}{125}\right) -\left(\frac{12}{5}\right)}$

$=\frac{\frac{9( 25) -108}{125}}{\frac{256-12( 25)}{125}}$

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

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

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

Let us consider RHS,

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

$=\frac{\frac{9}{4} -\left(\frac{27}{64}\right)}{1-\left(\frac{27}{16}\right)}$

$=\frac{\frac{9( 16) -27}{64}}{\frac{16-27}{16}}$

$=\frac{144-27}{4( -11)}$

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

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

LHS $=$ RHS

Hence proved.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

21 Views