if tan alpha =1/3 and tan beta =1/7 prove that tan(2 alpha+beta)=45 degrees
tan ( A + B ) = ( tan A + tan B ) / ( 1 - tan A * tan B )
In this case:
A = 2 alpha , B = beta
so
tan ( 2 alpha + beta ) = [ tan ( 2 alpha ) + tan ( beta ) ] / [ 1 - tan ( 2 alpha ) * tan ( beta ) ]
Since the:
tan ( 2 alpha ) = 2 * tan ( alha ) / [ 1 - tan ^ 2 ( alpha ) ]
put:
tan ( alpha ) = 1 / 3 in this formula
tan ( 2 alpha ) = 2 * ( 1 / 3) / [ 1 - ( 1 / 3) ^ 2 ]
tan ( 2 alpha ) = ( 2 / 3 ) / ( 1 - 1 / 9 )
tan ( 2 alpha ) = ( 2 / 3 ) / ( 9 / 9 - 1 / 9 )
tan ( 2 alpha ) = ( 2 / 3 ) / ( 8 / 9 )
tan ( 2 alpha ) = ( 2 * 9 ) / ( 8 * 3 )
tan ( 2 alpha ) = 18 / 24
tan ( 2 alpha ) = 6 * 3 / ( 6 * 4 )
tan ( 2 alpha ) = 3 / 4
Replace:
tan ( 2 alpha ) = 3 / 4 and tan ( beta ) = 1 / 7 in formula
tan ( 2 alpha + beta ) = [ tan ( 2 alpha ) + tan ( beta ) ] / [ 1 - tan ( 2 alpha ) * tan ( beta ) ]
tan ( 2 alpha + beta ) = ( 3 / 4 + 1 / 7 ) / [ 1 - ( 3 / 4 ) * ( 1 / 7 ) ]
tan ( 2 alpha + beta ) = [ 3 * 7 / ( 4 * 7 ) + 1 * 4 / ( 7 * 4 ) ] / ( 1 - 3 / 28 )
tan ( 2 alpha + beta ) = ( 21 / 28 + 4 / 28 ) / [ 28 / 28 - 3 / 28 ) ]
tan ( 2 alpha + beta ) = ( 25 / 28 ) / (25 / 28 )
tan ( 2 alpha + beta ) = 1
tan ( 2 alpha + beta ) = tan ( 45 ° )
becouse:
tan ( 45 ° ) = 1
So:
tan ( 2 alpha + beta ) = tan ( 45 ° )
2 alpha + beta = 45 °
Helpful answer
To prove that tan(2α + β) equals 45 degrees, we will start by using trigonometric identities to express tan(2α + β) in terms of tan α and tan β.
Step 1: Recall the double angle identity for tangent:
tan(2θ) = (2tanθ) / (1 - tan²θ)
Step 2: Apply the double angle identity to tan(2α):
tan(2α) = (2tanα) / (1 - tan²α)
Step 3: Substitute the given values of tan α and tan β into the equations:
tan α = 1/3
tan β = 1/7
tan(2α) = (2(1/3)) / (1 - (1/3)²)
= (2/3) / (1 - 1/9)
= (2/3) / (8/9)
= 3/4
Step 4: Now, let's express tan(2α + β) in terms of tan(2α) and tan β:
tan(2α + β) = (tan(2α) + tan β) / (1 - tan(2α)tan β)
Step 5: Substitute the values of tan(2α) and tan β:
tan(2α + β) = (3/4 + 1/7) / (1 - (3/4)(1/7))
Step 6: Perform the calculations:
tan(2α + β) = (21/28 + 4/28) / (1 - 3/28)
= (25/28) / (25/28)
= 1
Step 7: Since tan(2α + β) = 1, we can conclude that 2α + β = 45 degrees.
Therefore, we have proven that tan(2α + β) is equal to 45 degrees.
To prove that tan(2α + β) = 45 degrees, we will use the trigonometric identities and the given information that tan(α) = 1/3 and tan(β) = 1/7.
First, let's start by using the double-angle formula for tangent:
tan(2θ) = (2tanθ) / (1 - tan²θ)
Therefore, tan(2α + β) = (2tan(α + β)) / (1 - tan²(α + β))
Using the addition formula for tangent:
tan(α + β) = (tanα + tanβ) / (1 - tanαtanβ)
We substitute the given values into the formula:
tan(α + β) = (1/3 + 1/7) / (1 - (1/3)(1/7))
Simplifying the expression:
tan(α + β) = (7/21 + 3/21) / (1 - (1/21))
tan(α + β) = 10/21 / (20/21)
tan(α + β) = 10/20
tan(α + β) = 1/2
Now, substitute the result back into the initial equation:
tan(2α + β) = (2tan(α + β)) / (1 - tan²(α + β))
tan(2α + β) = (2 * 1/2) / (1 - (1/2)²)
tan(2α + β) = 1 / (1 - 1/4)
tan(2α + β) = 1 / (3/4)
tan(2α + β) = 4/3
To determine the angle in degrees where tan(2α + β) = 4/3, we can use the inverse tangent function:
2α + β = arctan(4/3)
2α + β ≈ 53.13 degrees
Therefore, tan(2α + β) is approximately equal to 53.13 degrees, not 45 degrees.
Hence, we have proven that, based on the given information, tan(2α + β) is not equal to 45 degrees.