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SINA=nSINB THEN n-1/n+1 tan A+B/2=______________ A]SIN(A-B/2) B] TAN (A-B/2) C] COT (A-B/2) D]NOT

SINA=nSINB THEN n-1/n+1 tan A+B/2=______________
A]SIN(A-B/2)
B] TAN (A-B/2)
C] COT (A-B/2)
D]NOT
 
 

Grade:11

1 Answers

Arun
25763 Points
3 years ago
we have sinA = k sinB.

implies k/1 = sinA /sinB.

applying dividendo and componendo ,we get
k-1/k+1=  sinA - sin B / sin A + sin B
=2 cos(A+B) / 2sin(A-B)/2 / 2  sin(A+B)  /  2 cos(A-B)/2

now finally we will have

tan(A-B)/2/ tan(A+B)/2  = k-1/k +1.


we obtain tan(A-B)/2
 
Option B is correct
 
 
Regards
Arun (askIITians forum expert)

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