Question

If 4 tanB=3 then 4sinB-3cosB/4sinB+3cosB

Answer 1

Answer:

0

Step-by-step explanation:

$\tt 4tan(B)=3$

$\tt tan(B)=\dfrac{3}{4}$

$\tt tan(B)=\dfrac{opposite\:side}{adjecent\:side}$

$\sf opposite\:side=3$

$\sf adjecent\:side=4$

By using Pythagoras theorem,

$\tt (a)^2+(b)^2=(c)^2$

$\tt (3)^2+(4)^2=(c)^2$

$\tt c=\sqrt{9+16}$

$\tt c=\sqrt{25}$

$\tt c=5$

$\sf sin(B)=\dfrac{opposite\:side}{hypotenuse}$

$\sf sin(B)=\dfrac{3}{5}$

$\sf cos(B)=\dfrac{adjecent\:side}{hypotenuse}$

$\sf cos(B)=\dfrac{4}{5}$

$\tt \dfrac{4sin(B)-3cos(B)}{4sin(B)+3cos(B)}$

$\tt =\dfrac{4\bigg(\dfrac{3}{5}\bigg)-3\bigg(\dfrac{4}{5}\bigg)}{4\bigg(\dfrac{3}{5}\bigg)+3\bigg(\dfrac{4}{5}\bigg)}$

$\tt =\dfrac{\dfrac{12}{5}-\dfrac{12}{5}}{\dfrac{12}{5}+\dfrac{12}{5}}$

$\tt= \dfrac{\dfrac{(60-60)}{5}}{\dfrac{(60+60)}{5}}$

$\tt= \dfrac{5(0)}{5(120)}$

$\tt= \dfrac{0}{600}$

$\tt= 0$