Question

27: if sec A = 5/4 then the value of 1
- tan A/1 + tan A is​

Answer 1

$secA = \frac{5}{4} \\to\:find \\\frac{1-tanA}{1+tanA} \\\\we \:know\:that\: sec \:theta= \frac{H}{B} \\so,\: hypotaneous = 5 \\ and\: base= 4 \\\\so \:by\:pythagoreus \: theorem \\perpendicular = 3 \\\\tan \:theta = \frac{P}{B} \\\\\frac{1-tanA}{1+tanA} \\=> \frac{1-\frac{3}{5} }{1+\frac{3}{5} } \\=> \frac{\frac{5-3}{5} }{\frac{5+3}{5} } \\=>\frac{2}{8} \\=> \boxed{\frac{1}{4} }$