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chaina

· started a discussion

· 1 Months ago

solution (Question: 115583.0 )

samaj ni aya

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,
In the ratio of Sin (A-B) / sin (A+B) , we have multiplied sin(A+B) in both numerator and denominator.
Now numerator becomes sin(A-B) sin(A+B) i.e = sin^2 A - sin^2 B ( by trignometric formula)
And in denominator we get sin^2(A+B) which can be written as sin^2(pie - C) , because in a triangle A+B+C= pie
Now you can solve the question further .

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