(B) We know that in $\triangle ABC$,$A+B+C = \pi$,so $A+B = \pi-C$. Substituting this into the expression: $2ab \sin \frac{1}{2}(A+B-C) = 2ab \sin \fr

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(B) We know that in $\triangle ABC$,$A+B+C = \pi$,so $A+B = \pi-C$. Substituting this into the expression: $2ab \sin \frac{1}{2}(A+B-C) = 2ab \sin \frac{1}{2}((\pi-C)-C)$ $= 2ab \sin \frac{1}{2}(\pi-2C) = 2ab \sin (\frac{\pi}{2}-C)$ $= 2ab \cos C$ Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab}$. Therefore,$2ab \cos C = 2ab \left(\frac{a^2+b^2-c^2}{2ab}\right) = a^2+b^2-c^2$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Medium

In $\triangle ABC$,with usual notations,$2ab \sin \frac{1}{2}(A+B-C) =$

MHT CET 2021 All MHT CET

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

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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