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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.
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