If A lies in third quadrant, and $20 \tan A = 21$, then the value of $\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}$.

1

20 tanA = 21

tanA = \(\frac{21}{20}\)

{ we know, tanA = \(\frac{P}{B}\) }

Now,

 \(\frac{5sinA - 2cosA }{4 cosA - 5/7sinA}\)

=  \(\frac{5 × P/H - 2 × B/H }{4  × B/H - 5/7 × P/H}\)

=  \(\frac{5 × P - 2 × B }{4  × B - 5/7 × P}\)

=  \(\frac{5 × 21 - 2 × 20 }{4  × 20 - 5/7 × 21}\)

= \(\frac{105 - 40 }{80 - 15}\) 

= \(\frac{65 }{65}\)

= 1