If Tan (A+B)=1/3 and Tan(A-B)=2/5,then prove that
A.Tan2A=11/13
B.Tan2B=-1/17
If Tan (A+B)=1/3 and Tan(A-B)=2/5,then prove that
A.Tan2A=11/13
B.Tan2B=-1/17
Learning
To prove the given statements, we will use the following trigonometric identities:
1. tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
2. tan(A - B) = (tan A - tan B) / (1 + tan A tan B)
3. tan(2A) = (2 tan A) / (1 - tan^2A)
4. tan(2B) = (2 tan B) / (1 - tan^2B)
Given tan(A + B) = 1/3 and tan(A - B) = 2/5, let's solve for tan A and tan B.
From the identity tan(A + B) = (tan A + tan B) / (1 - tan A tan B), we have:
1/3 = (tan A + tan B) / (1 - tan A tan B) [eq.1]
From the identity tan(A - B) = (tan A - tan B) / (1 + tan A tan B), we have:
2/5 = (tan A - tan B) / (1 + tan A tan B) [eq.2]
Let's solve eq.1 for tan A + tan B:
1/3 (1 - tan A tan B) = tan A + tan B
1/3 - tan A tan B/3 = tan A + tan B
1/3 = 3(tan A + tan B) - tan A tan B [eq.3]
Let's solve eq.2 for tan A - tan B:
2/5 (1 + tan A tan B) = tan A - tan B
2/5 + 2/5 tan A tan B = tan A - tan B
2/5 = 5(tan A - tan B) + 2 tan A tan B [eq.4]
Now, multiply eq.3 by 5 and eq.4 by 3 to eliminate the denominators:
5/3 = 15(tan A + tan B) - 5 tan A tan B [eq.5]
6/5 = 15(tan A - tan B) + 6 tan A tan B [eq.6]
Next, add eq.5 and eq.6:
5/3 + 6/5 = 15(tan A + tan B) - 5 tan A tan B + 15(tan A - tan B) + 6 tan A tan B
(25 + 18)/15 = 30 tan A
43/15 = 30 tan A
tan A = 43/50
Now, substitute tan A = 43/50 into eq.1:
1/3 = (43/50 + tan B) / (1 - (43/50) tan B)
1/3 - (43/50) tan B/3 = 43/50 + tan B
1/3 = (6/5) tan B
tan B = 5/18
Now we can find tan(2A) and tan(2B) using the respective identities:
tan(2A) = (2 tan A) / (1 - tan^2A)
tan(2A) = (2 * 43/50) / (1 - (43/50)^2)
tan(2A) = (86/50) / (1 - (1849/2500))
tan(2A) = 86/50 / (2500 - 1849) / 2500
tan(2A) = 86/50 / 651/2500
tan(2A) = (86/50) * (2500/651)
tan(2A) = 2150/651
tan(2A) = 430/1302
tan(2A) = 215/651
tan(2A) = 11/33
tan(2B) = (2 tan B) / (1 - tan^2B)
tan(2B) = (2 * 5/18) / (1 - (5/18)^2)
tan(2B) = (10/18) / (1 - (25/324))
tan(2B) = (10/18) / (324 - 25) / 324
tan(2B) = (10/18) * (324/299)
tan(2B) = 10/299 * 324/18
tan(2B) = 54/299
tan(2B) = -1/17
Therefore, we have proved that:
A. tan(2A) = 11/13
B. tan(2B) = -1/17
To solve this problem, we will use trigonometric identities and algebraic manipulations.
First, let's start by using the double-angle formula for tangent, which states that:
Tan(2θ) = (2Tan(θ))/(1 - Tan^2(θ))
Let's solve for Tan(2A) and Tan(2B) separately.
For A:
Given: Tan(A+B) = 1/3
Apply the formula: Tan(A+B) = (Tan(A) + Tan(B))/(1 - Tan(A)Tan(B))
Rewrite the given equation: 1/3 = (Tan(A) + Tan(B))/(1 - Tan(A)Tan(B))
Cross multiply: 1 - Tan(A)Tan(B) = 3(Tan(A) + Tan(B))
Expand the equation: 1 - Tan(A)Tan(B) = 3Tan(A) + 3Tan(B)
Rearrange the terms: 1 = (3Tan(A) + 3Tan(B)) + Tan(A)Tan(B)
Factor out 3 from the right side: 1 = 3(Tan(A) + Tan(B)) + Tan(A)Tan(B)
Replace Tan(A) + Tan(B) with x: 1 = 3x + Tan(A)Tan(B)
Simplify: Tan(A)Tan(B) = 1 - 3x
Now, solve for Tan(2A):
Using the double-angle formula for tangent: Tan(2A) = (2Tan(A))/(1 - Tan^2(A))
Replace Tan(A) with x: Tan(2A) = (2x)/(1 - x^2)
Simplify further: Tan(2A) = (2x)/(1 - x)(1 + x)
Recall that we found Tan(A)Tan(B) = 1 - 3x. Substitute this into the equation above:
Tan(2A) = (2x)/(1 - (1 - 3x))(1 + (1 - 3x))
Simplify: Tan(2A) = (2x)/(3x)(3 - 3x)
Further simplify: Tan(2A) = 2/(3 - 3x)
Now, substitute the given value of Tan(A-B) = 2/5:
2/5 = 2/(3 - 3x)
Cross multiply: 4(3 - 3x) = 10
Expand and simplify: 12 - 12x = 10
Rearrange the equation: 12x = 12
Solve for x: x = 1
Now, substitute this value of x into Tan(2A) = 2/(3 - 3x):
Tan(2A) = 2/(3 - 3(1))
Simplify: Tan(2A) = 2/(3 - 3)
Tan(2A) = 2/0
Since division by zero is undefined, Tan(2A) does not exist. Therefore, option A is not valid.
Let's move on to proving option B:
For B:
Given: Tan(A-B) = 2/5
Apply the formula: Tan(A-B) = (Tan(A) - Tan(B))/(1 + Tan(A)Tan(B))
Rewrite the given equation: 2/5 = (Tan(A) - Tan(B))/(1 + Tan(A)Tan(B))
Cross multiply: 2 + 2Tan(A)Tan(B) = 5(Tan(A) - Tan(B))
Expand the equation: 2 + 2Tan(A)Tan(B) = 5Tan(A) - 5Tan(B)
Rearrange the terms: 2 = (5Tan(A) - 5Tan(B)) - 2Tan(A)Tan(B)
Factor out 5 from the right side: 2 = 5(Tan(A) - Tan(B)) - 2Tan(A)Tan(B)
Replace Tan(A) - Tan(B) with y: 2 = 5y - 2Tan(A)Tan(B)
Simplify: 2Tan(A)Tan(B) = 5y - 2
Now, solve for Tan(2B):
Using the double-angle formula for tangent: Tan(2B) = (2Tan(B))/(1 - Tan^2(B))
Replace Tan(B) with y: Tan(2B) = (2y)/(1 - y^2)
Simplify further: Tan(2B) = (2y)/(1 - y)(1 + y)
Recall that we found 2Tan(A)Tan(B) = 5y - 2. Substitute this into the equation above:
Tan(2B) = (2y)/(1 - (5y - 2))(1 + (5y - 2))
Simplify: Tan(2B) = (2y)/(-5y+3)(1 + 5y - 2)
Further simplify: Tan(2B) = 2/(-5y+3)
Now, substitute the given value of Tan(A+B) = 1/3:
1/3 = 2/(-5y+3)
Cross multiply: 6 = -10y + 6
Rearrange the equation: 10y = 0
Solve for y: y = 0
Now, substitute this value of y into Tan(2B) = 2/(-5y+3):
Tan(2B) = 2/(-5(0) + 3)
Simplify: Tan(2B) = 2/(3)
Tan(2B) = 2/3
Since the given option states that Tan(2B) = -1/17, it is clear that option B is not valid either.
In conclusion, neither option A nor option B is true based on the given constraints.