In ∆ABC,(a-b)2cos2C2+(a+b)2sin2C2 is equal to
a2
b2
c2
None of these
Explanation for the correct option:
Solving the expression:
Given
(a-b)2cos2C2+(a+b)2sin2C2=a2+b2-2abcos2C2+a2+b2+2absin2C2=(a2+b2)cos2C2+sin2C2-2abcos2C2-sin2C2
Using the identity sin2A+cos2A=1.
=(a2+b2)-2abcos2C2-sin2C2
Now, cos2A-sin2A=cos2A
⇒cos2C2-sin2C2=cos2C2⇒cos2C2-sin2C2=cosC
∴(a-b)2cos2C2+(a+b)2sin2C2=(a2+b2)-2abcosC
We know in ∆ABC
cosC=a2+b2-c22ab
Substituting the value of cosC in the above equation, we get
∴(a-b)2cos2C2+(a+b)2sin2C2=(a2+b2)-2abcos2C2-sin2C2=(a2+b2)-2abcosC=(a2+b2)-2aba2+b2-c22ab=a2+b2-a2-b2+c2=c2
Hence, option (C) is the correct answer.
In a triangle ABC, 2ca sinA−B+C2 is equal to
In a △ABC,2casin(A−B+C2) is equal to