Question

if sec theta =13 upon 5 show that (2 sin theta-3cos theta ) upon (4 sin theta -9 cos theta ) =3

Answer 1

let theta=x

secx=13/5

cosx=5/13

sinx12/13(pythagores theorem)

given

=2sinx-3cosx/4sinx-9cosx

=

$\frac{2 \times \frac{12}{13} - 3 \times \ \frac{5}{13} }{4 \times \frac{12}{13} - 9 \times \frac{5}{13} }$

=

$\frac{ \frac{9}{13} }{ \frac{3}{13} }$

=3

hence proved

Answer 2

Hey mate here we go

$given \: sec \: theta \: = \frac{13}{5} \\ \: but \: we \: know \: that \: \\ \cos \: theta \: = \frac{1}{ \sec \: theta } \\ cos \: theta = \frac{1}{ \frac{13}{5} } = \frac{5}{13} \\ we \: can \: find \: sin \: theta \: by \\ \: pythogarus \: theorem \\ sin \: theta \: = \frac{12}{13} \\ \\ nw \: given \: problem \: is \: that \\ 3 = \frac{2sint \: theta - 3cos \: theta}{4sin \: theta - 9cos \: theta} \\ 3= \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13} }{4 \times \frac{12 }{13} - 9 \times \frac{5}{13} } \\ 3 = \frac{ \frac{24}{13} - \frac{15}{13} }{ \frac{48}{13} - \frac{45}{13} } \\ 3 = \frac{ \frac{9}{13} }{ \frac{3}{13} } \\ 3= \frac{9}{13} \times \frac{13}{3} \\ 3 = 3 \\ \\ hence \: proved \: ur \: ans$

I hope this will helps you

Mark my ans as brainliest