Exams › JEE Advanced › Maths
If A + B + C = 180 deg, then the value of (sin2A + sin2B + sin2C)/(sinA + sinB + sinC) is:
- 8*sin(A/2)*sin(B/2)*sin(C/2)
- 4*cos(A/2)*cos(B/2)*cos(C/2)
- 8*cos(A/2)*cos(B/2)*cos(C/2)
- 4*sin(A/2)*sin(B/2)*sin(C/2)
Correct answer: 8*sin(A/2)*sin(B/2)*sin(C/2)
Solution
Using the two standard triangle identities for the numerator and denominator, the ratio simplifies to 8*sin(A/2)*sin(B/2)*sin(C/2).
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