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If tanA = n / n+1 , and tanB = 1 / 2n+1 then find (A+B)?
Find the solution for the above trigo question please.
3 Answers
- Mein Hoon NaLv 76 years agoFavorite Answer
tan A = n/(n+1)
tan B= 1/(2n+1)
tan (A+B) = (tan A + tan B)/(1 - tan A + tan B) = ( n/(n+1) - 1/(2n+1)/( 1- n/(n+1) * 1/(2n+1))
= ( n(2n + 1) - (n+1))/((n+1)(2n+1) - (n+ 1)) = 1
as A and B both < pi/2 so so A + B < pi so A + B = pi/4
- DWReadLv 76 years ago
Needs parentheses!
tanA = n/(n+1)
tanB = 1/(2n+1)
A+B = arctan(n/(n+1)) + arctan(n/(2n+1))
- 6 years ago
tan(a) = n / (n + 1)
tan(b) = 1 / (2n + 1)
a = arctan(n / (n + 1))
b = arctan(1 / (2n + 1))
a + b = arctan(n / (n + 1)) + arctan(1 / (2n + 1))
tan(a + b) = tan(arctan(n / (n + 1)) + arctan(1 / (2n + 1))
tan(a + b) = (tan(arctan(n / (n + 1)) + tan(arctan(1 / (2n + 1))) / (1 - tan(arctan(n / (n + 1)) * tan(arctan(1 / (2n + 1)))
tan(a + b) = (n/(n + 1) + 1/(2n + 1)) / (1 - n * 1 / ((n + 1) * (2n + 1)))
tan(a + b) = ((n * (2n + 1) + 1 * (n + 1)) / ((n + 1) * (2n + 1) - n)
tan(a + b) = (2n^2 + n + n + 1) / (2n^2 + n + 2n + 1 - n)
tan(a + b) = (2n^2 + 2n + 1) / (2n^2 + 2n + 1)
tan(a + b) = 1
a + b = arctan(1)
a + b = pi/4
