With usual notations- is a ABC b 2- c 2- a c - c 2- a 2- b c - a 2- b 2- c c is equal toA- 1B- 0C- abcD- None of these

The correct option is B 0 We have b^2 c^2/a.cos A + c^2 a^2/b.cos B + a^2 b^2/c.cos C =b^2 c^2/a.b^2+c^2 a^2/2bc + c^2 a^2/b.c^2+a^2 b^2/2ac+a^2 b^2/b.c^2+a^2 ...

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Standard XI

Mathematics

A.G.P

With usual no...

Question

With usual notations, is a $△ A B C \frac{b^{2} - c^{2}}{a s e c c} + \frac{c^{2} - a^{2}}{b s e c c} + \frac{a^{2} - b^{2}}{c s e c c}$ is equal to

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abc

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None of these

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Solution

The correct option is B
0

We have
$\frac{b^{2} - c^{2}}{a} . c o s A + \frac{c^{2} - a^{2}}{b} . c o s B + \frac{a^{2} - b^{2}}{c} . c o s C$
= $\frac{b^{2} - c^{2}}{a} . \frac{b^{2} + c^{2} - a^{2}}{2 b c} + \frac{c^{2} - a^{2}}{b} . \frac{c^{2} + a^{2} - b^{2}}{2 a c} + \frac{a^{2} - b^{2}}{b} . \frac{c^{2} + a^{2} - b^{2}}{2 a c}$
=0

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With usual notations, is a △ABC b2−c2a sec c + c2−a2b sec c + a2−b2c sec c is equal to

The correct option is B 0 We have b2−c2a.cosA + c2−a2b.cosB + a2−b2c.cosC =b2−c2a.b2+c2−a22bc + c2−a2b.c2+a2−b22ac+a2−b2b.c2+a2−b22ac =0

With usual notations, is a $△ A B C \frac{b^{2} - c^{2}}{a s e c c} + \frac{c^{2} - a^{2}}{b s e c c} + \frac{a^{2} - b^{2}}{c s e c c}$ is equal to