In ABC- Prove that 2acsinA-B-C2-a2-c2-b2

We have A B+C2 =A+C B2 =π B B2 #160; #160; #160; #160; since A+B+C=π⇒ A+C=π B =π 2B2 =π2 B ∴ 2acsin(A B+C2) ...

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General Solution of Trigonometric Equation

In ABC, Pro...

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In $△ A B C,$ Prove that $2 a c sin (\frac{A - B + C}{2}) = a^{2} + c^{2} - b^{2}$

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> 0

= 8

a^{2} + b^{2} + c^{2} \geq 12

a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

Mark the correct alternative in each of the following :

$I n a n y Δ A B C, the value of 2ac s i n (\frac{A - B + C}{2}) i s$

A B C

\frac{tan A}{tan B} = \frac{c^{2} + a^{2} - b^{2}}{c^{2} + b^{2} - a^{2}}

2 a c \sin (\frac{A - B + C}{2})

a^{2} + b^{2} - c^{2}

c^{2} + a^{2} - b^{2}

b^{2} - c^{2} - a^{2}

c^{2} - a^{2} - b^{2}

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