Prove that
tan-1 (1/7) + tan-1 (1/13) = tan-1(2/9)
let tanA = 1/7 and tanB = 1/13
then tan(A+B) = (tanA + tanB)/( 1 - tanAtanB)
= (1/7 + 1/13)/(1 - (1/7)(1/13) )
= (20/91) / (90/91) = 20/90 = 2/9
and if we let tanC = 2/9
then A + B = C
thus tan^-1 (1/7) + tan^-1 (1/13) = tan^-1 (2/9)
check with your calculator
take 2nd Tan (1/7) = 8.13010..
take 2nd Tan(1/13) = 4.39871
add them: 12.5288..
take 2nd Tan(2/9) = 12.5288..
To prove the given equation:
Step 1: Start by expressing all the tangent inverse functions in terms of tangent.
Using the relationship between the tangent and its inverse function, we have:
tan(tan^(-1)(x)) = x
Therefore, we can write:
tan(tan^(-1)(1/7)) = 1/7
tan(tan^(-1)(1/13)) = 1/13
tan(tan^(-1)(2/9)) = 2/9
Step 2: Use the addition formula for tangent.
The addition formula for tangent states that:
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)*tan(b))
In this case, let a = tan^(-1)(1/7) and b = tan^(-1)(1/13).
Now, substitute the expressions for a and b into the addition formula:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = (tan(tan^(-1)(1/7)) + tan(tan^(-1)(1/13))) / (1 - tan(tan^(-1)(1/7))*tan(tan^(-1)(1/13)))
Simplifying this equation, we get:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = (1/7 + 1/13) / (1 - (1/7)*(1/13))
Step 3: Simplify the equation.
To simplify, find a common denominator for the fractions in the numerator:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = (13/91 + 7/91) / (1 - 1/91)
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = (20/91) / (90/91)
Now, divide the numerator by the denominator:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = (20/91) * (91/90)
Simplifying this expression:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = 20/90
Reducing the fraction gives:
tan(tan^(-1)(1/7) + tan^(-1)(1/13)) = 2/9
Thus, we have shown that:
tan^(-1)(1/7) + tan^(-1)(1/13) = tan^(-1)(2/9)