Question

if sec theta = 13/5, show that 2sintheta - 3costheta/4sintheta - 9costheta = 3

Answer 1

$= > sec \: a \: = \frac{13}{5} \\ \\ = > \frac{1}{cos \: a} = \frac{5}{13} \\ \\ sin \: a = \sqrt{1 - {( \frac{5}{13} )}^{2} } \\ \\ = \sqrt{ \frac{144}{169} } \\ \\ = \frac{12}{13} \\ \\ tan \: a \: = \frac{sin \: a}{cos \: a} = > \frac{12}{5} \\ \\ \\ taking \: lhs \\ = \frac{2sin \: a \: - 3 \: cos \: a}{4sin \: a - 9cos \: a} \\ \\ = \frac{2tan \: a - 3}{4tan \: a - 9} \:\:(divide\: by\: cos\:a) \\ \\ = \frac{2 \times \frac{12}{5} - 3 }{4 \times \frac{12}{5} - 9} \\ \\ = \frac{24 - 15}{48 - 45} \\ \\ = \frac{9}{3} \\ \\ = 3 \\ \\= rhs$