if tan(a+b)=2/3 and tan(a-b)=2/5 then find the value of tan2b.
If you want just numerical value, then
a+b=atan(2/3)=33.69°
a-b=atan(2/5)=21.80°
2b=atan(2/3)-atan(2/5)
tan(2b)=tan(atan(2/3)-atan(2/5))
=0.210526
tan(atan(2/3)-atan(2/5))
= (2/3 - 2/5)/(1+(2/3)(2/5))
= 4/19
= 0.21
as above
To find the value of tan(2b), we first need to manipulate the given information to express it in terms of tan(2b). Let's start by using a trigonometric identity.
The identity we will use is:
tan(2θ) = (2tan(θ))/(1-tan^2(θ))
Let's solve for tan(b) using the given equations, tan(a+b) = 2/3 and tan(a-b) = 2/5:
tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b))
2/3 = (tan(a)+tan(b))/(1-tan(a)tan(b)) ----(1)
tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b))
2/5 = (tan(a)-tan(b))/(1+tan(a)tan(b)) ----(2)
Now, let's manipulate equations (1) and (2) to solve for tan(a) and tan(b):
From equation (1):
2/3 = (tan(a)+tan(b))/(1-tan(a)tan(b))
2/3(1-tan(a)tan(b)) = tan(a)+tan(b)
2(1-tan(a)tan(b)) = 3(tan(a)+tan(b))
2 - 2tan(a)tan(b) = 3tan(a) + 3tan(b)
2 - 2tan(a)tan(b) - 3tan(a) - 3tan(b) = 0
2 - 3tan(a) - 3tan(b) - 2tan(a)tan(b) = 0
2 - (3tan(a) + 3tan(b) + 2tan(a)tan(b)) = 0
Similarly, from equation (2):
2 - (3tan(a) - 3tan(b) - 2tan(a)tan(b)) = 0
Now, we have two equations with two variables (tan(a) and tan(b)). If we solve these two equations simultaneously, we can find the values of tan(a) and tan(b). Since we only need to find the value of tan(2b), we can substitute back into the initial equation:
tan(2b) = (2tan(b))/(1 - tan^2(b))