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Laxmikant

· started a discussion

· 1 Months ago

nice question (Question: 115583.0 )

mstttttttttt

Question:
If any \(\triangle\)ABC, if a, b and c are the sides opposite to angles A, B and C respectively, then \(\cfrac{sin(A-B)}{sin(A+B)}\)   is equal to :
Options:
A) \(\cfrac{a^2-b^2}{c^2}\)
B) \(\cfrac{b^2-a^2}{c^2}\)
C) \(\cfrac{c^2-b^2}{a^2}\)
D) \(\cfrac{b^2-c^2}{a^2}\)
Solution:
Ans: (a)

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Thanks for your support.

Thanks and Regards
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