Given that tan 75 degrees=2+√3 without using a table or a calculator determine 5 tan 15 degrees leaving your answer in the form a+b√3
We know that tan 75 degrees = 2+√3.
Using the identity tan (A+B) = (tan A + tan B) / (1 - tan A tan B), we can rewrite 75 degrees as (45 degrees + 30 degrees).
tan (45 degrees + 30 degrees) = (tan 45 degrees + tan 30 degrees) / (1 - tan 45 degrees * tan 30 degrees).
We know that tan 45 degrees = 1, and tan 30 degrees = 1/√3.
Substituting these values, we get:
tan (45 degrees + 30 degrees) = (1 + 1/√3) / (1 - 1 * 1/√3).
Simplifying the expression, we can multiply the numerator and denominator by √3 to eliminate the square root from the denominator:
= (√3 + 1) / (√3 - 1/√3)
= (√3 + 1) / (√3 - 1/√3)
= (√3 + 1) / (√3 - 1/√3) * (√3/√3)
= (√3 + √3) / (√3*√3 - 1)
= (2√3) / (3-1)
= (2√3) / 2
= √3
So, tan (75 degrees) = √3.
We are required to find 5 tan 15 degrees.
Using the identity tan (A-B) = (tan A - tan B) / (1 + tan A tan B), we can rewrite 15 degrees as (45 degrees - 30 degrees).
tan (45 degrees - 30 degrees) = (tan 45 degrees - tan 30 degrees) / (1 + tan 45 degrees * tan 30 degrees).
We know that tan 45 degrees = 1, and tan 30 degrees = 1/√3.
Substituting these values, we get:
tan (45 degrees - 30 degrees) = (1 - 1/√3) / (1 + 1 * 1/√3).
Simplifying the expression, we can multiply the numerator and denominator by √3 to eliminate the square root from the denominator:
= (√3 - 1) / (√3 + 1/√3)
= (√3 - 1) / (√3 + 1/√3)
= (√3 - 1) / (√3 + 1/√3) * (√3/√3)
= (√3 - √3) / (√3*√3 + 1)
= (0) / (3 + 1)
= 0
Therefore, 5 tan 15 degrees = 5 * 0 = 0.
The answer is 0.
We can use the trigonometric identity for tan(2θ) to find tan 15 degrees.
The trigonometric identity for tan(2θ) is given by:
tan(2θ) = (2 * tan(θ))/(1 - tan^2(θ))
Since we know the value of tan 75 degrees, we can express it as tan(2θ) where θ = 37.5 degrees. Hence, tan(37.5 degrees) = (2 * tan(15 degrees))/(1 - tan^2(15 degrees))
Let's substitute tan(37.5 degrees) = 2 + √3 and solve for tan(15 degrees).
2 + √3 = (2 * tan(15 degrees))/(1 - tan^2(15 degrees))
Now, let's solve for tan(15 degrees):
(2 * tan(15 degrees))/(1 - tan^2(15 degrees)) = 2 + √3
Cross-multiplying:
2 * tan(15 degrees) = (2 + √3) * (1 - tan^2(15 degrees))
Expanding the right side:
2 * tan(15 degrees) = 2 + √3 - (2 * tan^2(15 degrees)) - (√3 * tan^2(15 degrees))
Combine like terms:
2 * tan(15 degrees) + (2 * tan^2(15 degrees)) + (√3 * tan^2(15 degrees)) = 2 + √3
Rearranging:
(2 * tan^2(15 degrees)) + (√3 * tan^2(15 degrees)) + 2 * tan(15 degrees) - (2 + √3) = 0
Now, we can factor out tan(15 degrees):
tan^2(15 degrees) * (2 + √3) + 2 * tan(15 degrees) * (1 - (2 + √3)) = 0
tan(15 degrees) * (tan(15 degrees) * (2 + √3) + 2 * (1 - (2 + √3))) = 0
Simplifying further:
tan(15 degrees) * (tan(15 degrees) * (2 + √3) - 2√3 - 2) = 0
Since we are looking for a non-zero solution, we can ignore the (tan(15 degrees) * (2 + √3) - 2√3 - 2) term.
Hence, tan(15 degrees) = 0
Therefore, 5 tan(15 degrees) = 5 * 0 = 0
So, the answer is 0.