In triangle ABC, sinA+sinB+sinCsinA+sinB−sinC=
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
tanA2cotB2
b
cotA2tanB2
c
cotA2cotB2
d
tanA2tanB2
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Denominator = sin A + sin B - sin C =2sinA+B2cosA−B2−2sinC2cosC2=2cosC2cosA−B2−sinC2=2cosC2cosA−B2−cosA+B2=2cosC22sinA2sinB2=4sinA2sinB2cosC2Also sinA+sinB+sinC=4cosA2cosB2cosC2⇒sinA+sinB+sinCsinA+sinB−sinC=cotA2cotB2