If $ \sec A=\frac{17}{8} $, verify that $ \frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A} $

Given:

$sec\ A = \frac{17}{8}$.

To do:

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

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $sec\ A=\frac{17}{8}$.

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,

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

$\Rightarrow (17)^2=(8)^2+BC^2$

$\Rightarrow BC^2=289-64$

$\Rightarrow BC=\sqrt{225}=15$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{15}{17}$

$cos\ \theta=\frac{AB}{AC}=\frac{8}{17}$

$tan\ \theta=\frac{BC}{AB}=\frac{15}{8}$

This implies,

Let us consider LHS,

$\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-4\left(\frac{15}{17}\right)^{2}}{4\left(\frac{8}{17}\right)^{2} -3}$

$=\frac{3-4\left(\frac{225}{289}\right)}{4\left(\frac{64}{289}\right) -3}$

$=\frac{\frac{3( 289) -4( 225)}{289}}{\frac{4( 64) -3( 289)}{289}}$

$=\frac{867-900}{256-867}$

$=\frac{-33}{-611}$

$=\frac{33}{611}$

Let us consider RHS,

$\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}=\frac{3-\left(\frac{15}{8}\right)^{2}}{1-3\left(\frac{15}{8}\right)^{2}}$

$=\frac{3-\left(\frac{225}{64}\right)}{1-3\left(\frac{225}{64}\right)}$

$=\frac{\frac{3( 64) -225}{64}}{\frac{64-3( 225)}{64}}$

$=\frac{192-225}{64-675}$

$=\frac{-33}{-611}$

$=\frac{33}{611}$

LHS $=$ RHS

Hence proved.