In a triangle ABC, 2ac sin12(A-B+C)=

In a triangle ABC, 2ac $\sin \frac{1}{2} (A - B + C) =$

A
${a^{2} + b^{2} - c^{2}}^{}$
B
$c^{2} + a^{2} - b^{2}$
C
$b^{2} - c^{2} - a^{2}$
D
$c^{2} - a^{2} - b^{2}$

Solution:

$\frac{1}{2} (A - B + C) = \frac{1}{2} (π - B - B) = \frac{π}{2} - B$

$∴ 2 a c$ $\sin \frac{1}{2} (A - B + C) = 2 a c \sin (\frac{π}{2} - B)$

= 2ac cos $B = a^{2} + c^{2} - b^{2}$ (by cosine rule)

Related content

JEE Main 2023 Session 2 Registration to begin today

JEE Main 2023 Result: Session 1

JEE Advanced 2023

NEET Rank Assurance Program | NEET Crash Course 2023

JEE Main 2023 Question Papers with Solutions

JEE Main 2024 Syllabus

Best Books for JEE Main 2024

JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria

JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria