In a △ ABC , 2 acsin 2 A − B + C is equal to, a 2 + b 2 − c 2, c 2 + a 2 − b 2, b 2 − a 2 − c 2, c 2 − a 2 − b 2

Hints : 2 ac sin ( 2 A + C − B ) [ 2 A + C = 2 π − 2 B ] , = 2 ac sin ( 2 π − B ) = 2 ac cos B = a 2 + c 2 − b 2

In a △ ABC , 2 acsin 2 A − B + C is equal to

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Question: Answered & Verified by Expert

In $\mathrm{a} \triangle \mathrm{ABC}, 2 \operatorname{acsin} \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}$ is equal to

MathematicsProperties of TrianglesWBJEEWBJEE 2010

Options:

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

Solution:

1526 Upvotes Verified Answer

The correct answer is: $c^2+a^2-b^2$

Hints : $2 \mathrm{ac} \sin \left(\frac{\mathrm{A}+\mathrm{C}-\mathrm{B}}{2}\right) \quad\left[\frac{\mathrm{A}+\mathrm{C}}{2}=\frac{\pi}{2}-\frac{\mathrm{B}}{2}\right],=2 \mathrm{ac} \sin \left(\frac{\pi}{2}-\mathrm{B}\right)=2 \mathrm{ac} \cos \mathrm{B} \quad=\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2$

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