If in a △ A BC , 2 b 2 = a 2 + c 2 , then s i n B s i n 3 B is equal to

s i n B s i n 3 B = s i n B 3 s i n B − 4 s i n 3 B = 3 − 4 sin 2 B = 3 − 4 ( 1 − cos 2 B ) = − 1 + 4 ( a c ) 2 4 ( a 2 + c 2 − b 2 ) 2 = − 1 + ( a c ) 2 ( 2 a 2 + c 2 ) 2 = − 1 + 4 ( a c ) 2 ( a 2 + c 2 ) 2 = ( 2 a c c 2 − a 2 ) 2

Mathematics - Properties of Triangles Question with Solution | TestHub

MathematicsProperties of TrianglesGeneralHard2 minPYQ_2022

MathematicsHardsingle choice

If in a $△ A BC, 2 b^{2} = a^{2} + c^{2}$ , then $\frac{s i n 3 B}{s i n B}$ is equal to

Options:

Answer:

Solution:

$\frac{s i n 3 B}{s i n B} = \frac{3 s i n B - 4 s i n ^{3} B}{s i n B}$ $= 3 - 4 sin^{2} B$ $= 3 - 4 (1 - cos^{2} B)$ $= - 1 + \frac{4 ( a ^{2} + c ^{2} - b ^{2} ) ^{2}}{4 ( a c ) ^{2}}$ $= - 1 + \frac{( \frac{a ^{2} + c ^{2}}{2} ) ^{2}}{( a c ) ^{2}}$ $= - 1 + \frac{( a ^{2} + c ^{2} ) ^{2}}{4 ( a c ) ^{2}}$ $= (\frac{c ^{2} - a ^{2}}{2 a c})^{2}$

Stream:BITSATSubject:MathematicsTopic:Properties of TrianglesSubtopic:General

⏱ 2mℹ️ Source: PYQ_2022

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