26. Let ABC be a triangle. If cos2A + cos2B+ cos2C= -1 then which one of the following is correct?
(a) sin A sin B sin C=0
(b) sin A sin B cos C= 0
(c) cos A sin B sin C = 0
(d) cos A cos B cos C = 0
Answers
Answer:
D
Step-by-step explanation:
"d" (and any subsequent words) was ignored because we limit queries to 32 words.
Given : ABC triangle.
cos2A + cos2B+ cos2C= -1
To Find : Choose correct option
Solution:
cos2A + cos2B+ cos2C= -1
=> cos2A + cos2B+ 1 + cos2C = 0
Using identity
cos x + cosy = 2 cos((x + y)/2)cos((x - y)/2)
1 + cos2x = 2cos²x
=> 2Cos(A + B) Cos(A - B) + 2Cos²C = 0
A + B = 180° - C ( angles of a triangle
cos (A + B) = cos (180° - C) = - cosC
=> 2(-CosC)Cos(A - B) + 2Cos²C = 0
=> -2Cosc ( Cos(A - B) - CosC) = 0
-CosC = Cos(A + B)
=>-2Cosc (cos(A - B) + cos(A + B) )= 0
=>-2Cosc ( cos(A + B) + Cos(A - B) ) = 0
=> -2Cosc (2cosA. CosB ) = 0
=> CosA.CosB.CosC = 0
Correct option is (d) cos A cos B cos C = 0
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