Inverse Trigonometric FunctionHard
Question
If sin-1a + sin-1b + sin-1c = π then the value of a
+ b 
+ c
=
+ b + c =Options
A.2abc
B.abc
C.1/2 abc
D.1/3 abc
Solution
sin-1a = A, sin-1b = B, sin-1c = C
A + B + C = π then
sin2A + sin2B + sin2C = 4sinA sin B sin C
sin A cos A + sin B cos B + sin C cos C = 2sin A sin B sin C
a
+ b 
+ c 
= 2abc
A + B + C = π then
sin2A + sin2B + sin2C = 4sinA sin B sin C
sin A cos A + sin B cos B + sin C cos C = 2sin A sin B sin C
a
+ b + c = 2abcCreate a free account to view solution
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