: If 4 tanβ = 3, then find the value of following expression. *
4 sinB -3cosB
4 sinB+3 cos B
Answers
Answer:
4tan(B)=3
\tt tan(B)=\dfrac{3}{4}tan(B)=
4
3
\tt tan(B)=\dfrac{opposite\:side}{adjecent\:side}tan(B)=
adjecentside
oppositeside
\sf opposite\:side=3oppositeside=3
\sf adjecent\:side=4adjecentside=4
By using Pythagoras theorem,
\tt (a)^2+(b)^2=(c)^2(a)
2
+(b)
2
=(c)
2
\tt (3)^2+(4)^2=(c)^2(3)
2
+(4)
2
=(c)
2
\tt c=\sqrt{9+16}c=
9+16
\tt c=\sqrt{25}c=
25
\tt c=5c=5
\sf sin(B)=\dfrac{opposite\:side}{hypotenuse}sin(B)=
hypotenuse
oppositeside
\sf sin(B)=\dfrac{3}{5}sin(B)=
5
3
\sf cos(B)=\dfrac{adjecent\:side}{hypotenuse}cos(B)=
hypotenuse
adjecentside
\sf cos(B)=\dfrac{4}{5}cos(B)=
5
4
\tt \dfrac{4sin(B)-3cos(B)}{4sin(B)+3cos(B)}
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)}=
4(
5
3
)+3(
5
4
)
4(
5
3
)−3(
5
4
)
\tt =\dfrac{\dfrac{12}{5}-\dfrac{12}{5}}{\dfrac{12}{5}+\dfrac{12}{5}}=
5
12
+
5
12
5
12
−
5
12
\tt= \dfrac{\dfrac{(60-60)}{5}}{\dfrac{(60+60)}{5}}=
5
(60+60)
5
(60−60)
\tt= \dfrac{5(0)}{5(120)}=
5(120)
5(0)
\tt= \dfrac{0}{600}=
600
0
\tt= 0=0