Question

In a ΔABC, if tan($\frac{A}{2}) = \frac{5}{6}$ and tan($\frac{B}{2}) = \frac{20}{37}$ then show that tan($\frac{C}{2}) = \frac{2}{5}$

Answer 1

in a ∆ABC, A + B + C = 180°

(A + B + C)/2 = 90°

tan(A + B + C)/2 = tan90°

or, (tanA/2 + tanB/2 + tanC/2 - tanA/2.tanB/2.tanC/2)/(1 - tanA/2.tanB/2 - tanB/2.tanC/2 - tanC/2.tanA/2) = 1/0 [ as we know, tan90° = ∞ = 1/0]

or, 1 = tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2


given, tanA/2 = 5/6, tanB/2 = 20/37


or, 1 = 5/6 × 20/37 + 20/37 × tanC/2 + tanC/2 × 5/6

or, 1 = 50/111 + 20tanC/2/37 + 5tanC/2/6

or, 1 - 50/111 = tanC/2 [ 20/37 + 5/6]

or, 61/111 = tanC/2 [ 305/222]

or, 61/111 = 305/222 × tanC/2

or, tanC/2 = 2/5 [ hence proved]

Answer 2

HELLO DEAR,



IN ∆ABC, A + B + C = 180°

(A + B + C)/2 = 90°

tan(A + B + C)/2 = tan90°

=> (tanA/2 + tanB/2 + tanC/2 - tanA/2.tanB/2.tanC/2)/(1 - tanA/2.tanB/2 - tanB/2.tanC/2 - tanC/2.tanA/2) = 1/0 [ as tan90° = ∞ = 1/0]

=> 1 = tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2

given:-
tanA/2 = 5/6, tanB/2 = 20/37


=> 1 = 5/6 × 20/37 + 20/37 × tanC/2 + tanC/2 × 5/6

=> 1 = 50/111 + 20tanC/2/37 + 5tanC/2/6

=> 1 - 50/111 = tanC/2 (20/37 + 5/6)

=> 61/111 = tanC/2 (305/222)

=> 61/111 = 305/222 × tanC/2

=> tanC/2 = 2/5


I HOPE IT'S HELP YOU DEAR,
THANKS