Sec=17/8 verify that (3-4sin²)/(4cos²-3)=(3-tan²)/(1-3tan²)

Given {tex}\sec A = \frac{{17}}{8} = \frac{{AC}}{{AB}}{/tex}
Let AC = 17K
and, AB = 8K
In {tex}\Delta ABC{/tex}, by Pythagoras theorem
BC² + AB² = AC²
BC² + (8K)² = (17K)²
BC² + 64K² = 289K²
BC² = 289K² - 64K²
BC² = 225K²
{tex}BC = \sqrt {225{K^2}} = 15K{/tex}
{tex}\therefore \sin A = \frac{{BC}}{{AC}} = \frac{{15K}}{{17K}} = \frac{{15}}{{17}}{/tex}
{tex}\cos A = \frac{{AB}}{{AC}} = \frac{{8K}}{{17K}} = \frac{8}{{17}}{/tex}
{tex}\tan A = \frac{{BC}}{{AB}} = \frac{{15K}}{{8K}} = \frac{{15}}{8}{/tex}
LHS
{tex} = \frac{{3 - 4{{\sin }^2}A}}{{4{{\cos }^2}A - 3}}{/tex}
{tex} = \frac{{3 - 4 \times {{\left( {\frac{{15}}{{17}}} \right)}^2}}}{{4{{\left( {\frac{8}{{17}}} \right)}^2} - 3}}{/tex}
{tex} = \frac{{3 - \frac{{900}}{{289}}}}{{\frac{{256}}{{289}} - 3}}{/tex}
{tex} = \frac{{\frac{{867 - 900}}{{289}}}}{{\frac{{256 - 867}}{{289}}}}{/tex}
{tex} = \frac{{\frac{{ - 33}}{{289}}}}{{\frac{{ - 611}}{{289}}}}{/tex}
{tex} = \frac{{ - 33}}{{289}} \times \frac{{289}}{{ - 611}}{/tex}
{tex} = \frac{{33}}{{611}}{/tex}
RHS
{tex} = \frac{{3 - {{\tan }^2}A}}{{1 - 3{{\tan }^2}A}}{/tex}
{tex} = \frac{{3 - {{\left( {\frac{{15}}{8}} \right)}^2}}}{{1 - 3 \times {{\left( {\frac{{15}}{8}} \right)}^2}}}{/tex}
{tex} = \frac{{3 - \frac{{225}}{{64}}}}{{1 - \frac{{675}}{{64}}}}{/tex}
{tex} = \frac{{\frac{{192 - 225}}{{64}}}}{{\frac{{64 - 675}}{{64}}}}{/tex}
{tex} = \frac{{\frac{{ - 33}}{{64}}}}{{\frac{{ - 611}}{{64}}}}{/tex}
{tex}= \frac{{ - 33}}{{64}} \times \frac{{64}}{{ - 611}}{/tex}
{tex} = \frac{{33}}{{611}}{/tex}
Hence verified