In - A B C - a b 2 - c 2 - c b 2 - a 2 - 0 then B-

The correct option is D π 3 [ a b^ 2 c^ 2 + c ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

In Δ A B C ...

Question

In $Δ A B C, \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0$ then $B =$

\frac{π}{2}

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\frac{π}{4}

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\frac{2 π}{3}

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\frac{π}{3}

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Solution

The correct option is D $\frac{π}{3}$
$\begin{matrix} \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0 \frac{sin A}{{sin}^{2} B - {sin}^{2} C} + \frac{sin C}{{sin}^{2} B - {sin}^{2} A} = 0 \frac{sin A}{sin (B + C) sin (B - C)} + \frac{sin C}{sin (B + A) sin (B - A)} = 0 \frac{1}{sin (B - C)} + \frac{1}{sin (B - A)} = 0 sin (B - A) + sin (B - C) = 0 2 sin (\frac{2 B - (A + C)}{2}) cos (\frac{C - A}{2}) = 0 2 B = A + C ∴ B = \frac{A + C}{2} A + B + C = π 3 B = π ∴ B = \frac{π}{3} H e n c e, o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

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In ΔABC,ab2−c2+cb2−a2=0 then B=

The correct option is D π3ab2−c2+cb2−a2=0sinAsin2B−sin2C+sinCsin2B−sin2A=0sinAsin(B+C)sin(B−C)+sinCsin(B+A)sin(B−A)=01sin(B−C)+1sin(B−A)=0sin(B−A)+sin(B−C)=02sin(2B−(A+C)2)cos(C−A2)=02B=A+C∴B=A+C2A+B+C=π3B=π∴B=π3Hence,optionDisthecorrectanswer.

In $Δ A B C, \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0$ then $B =$