Gujarat Boardગુજરાતી માધ્યમધોરણ 11 સાયન્સગણિતત્રિકોણમિતિય વિધયો1 Mark
MCQ
જો $\triangle ABC$ માં $sin^3+sin^3 B+sin^3 C=3 A sin B sin C, $
✓
0
B
1
C
3
D
5
✓
Answer
Correct option: A.
0
A
$\begin{vmatrix}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\b &c & a \\c & a & b
\end{vmatrix}=
\begin{vmatrix}
\mathbf{a+b+c} & \mathbf{b} & \mathbf{c} \\b+c+a &c & a \\c+a+b & a & b
\end{vmatrix}$
$\left[C_1\rightarrow C_1+C_2+C_3\right]$ કરતા
$=(a+b+c)\begin{vmatrix}
\mathbf{1} & \mathbf{b} & \mathbf{c} \\1 &c & a \\1 & a & b
\end{vmatrix}$
$=(a+b+c)\begin{vmatrix}
\mathbf{1} & \mathbf{b} & \mathbf{c} \\0 &c-a & a-c \\0 & a-c & b-a
\end{vmatrix}$
કરતા $\left[R_2\rightarrow R_2 -R_1 R_3\rightarrow R_3-R_2\right]$ કરતાપ
$=(a+b+c+)\left[(1)(c-b(c-b)(b-a)-(a-c))^2\right]$
$=(a+b+c)(bc-ac-b^2+ab-a^2-c^2+2ac)$
$=(a+b+c)(bc+ca+ab-a^2-b^2-b^2-c^2)$
$=-(a^3+b^3+c^3-3abc)$
$=-\left[(2R \ sin \ A)^3+(2R \ sin \ B)^3+(2R \ sin \ C^2)-3(2R \ sin \ A)(2R \ sin \ B)(2R \ sin \ C)\right]$
$=-8R(sin^3A+sin^3B+sin^3C-3sin \ A \ sin \ B \ sin \ C$
$=-8R^3(0)=0.$
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