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JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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Application +4 | -1
In a △ABC,2casin⁡(A−B+C2)\triangle A B C, 2 c a \sin \left(\frac{A-B+C}{2}\right) is equal to
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Correct option: B
c2+a2−b2c^2+a^2-b^2
Option B (c2+a2−b2c^2+a^2-b^2) is correct because by simplifying the trigonometric expression and using the Law of Cosines, we can show that 2casin⁡(A−B+C2)=c2+a2−b22 c a \sin \left(\frac{A-B+C}{2}\right) = c^2+a^2-b^2.

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