If 5 cot A = 3 then find the value of (5 sinA - 3 cosA)/(4 sinA + 3 CosA).

Question

If 5 cot A = 3 then find the value of \(\frac{5 sin A\, -\, 3 cos A}{4 sin A \,+\, 3 cos A}\).

Answer 1

Given,

5 Cot A = 3

Cot A = \(\frac{3}{5}\)

\(\frac{1}{tan A} = \frac{3}{5}\)

tan A = \(\frac{5}{3}\)

Now,

\(\frac{5sin A - 3cosA}{4sinA + 3cosA}\)

= \(\frac{5 \frac{sinA}{cos A}-3 \frac{cosA}{cos A}}{4 \frac{sinA}{cos A}+ 3 \frac{cosA}{cos A}}\) [dividing numerators and denominator by cos A]

= \(\frac{5tan A - 3}{4tan A + 3}\)

= \(\frac{5 \times \frac{5}{3} - 3}{4 \times \frac{5}{3}+3}\)

= \(\frac{\frac{25-9}{3}}{\frac{20+9}{3}}\)

= \(\frac{16}{29}\)