In any - ABC- prove that b2-c2-a2sin 2 A-c2-a2-b2sin 2 B-a2-b2-c2sin 2 C-0

By the sine rule, we have a/sin A + b/sin B + c/sin C = k(say) ⇒ a = k sin A, b = k sin B, and c = k sin C. Applying sine rule, asnd cosine formula, we get (b^ ...

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Byju's Answer

Standard VII

Mathematics

Equal Angles Subtend Equal Sides

In any Δ ABC,...

Question

In any $Δ A B C$ , prove that

$\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = 0$

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Solution

By the sine rule, we have

$\frac{a}{s i n A} + \frac{b}{s i n B} + \frac{c}{s i n C} = k (s a y)$

$\Rightarrow a = k s i n A, b = k s i n B, a n d c = k s i n C .$

Applying sine rule, asnd cosine formula, we get

$\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A = \frac{(b^{2} - c^{2})}{a^{2}} . (2 s i n A c o s a)$

= $\frac{(b^{2} - c^{2})}{a^{2}} (\frac{2 a}{k}) . \frac{(b^{2} + c^{2} - a^{2})}{2 b c}$

$[∴ s i n A = \frac{a}{k} a n d c o s A = \frac{(b^{2} + c^{2} - a^{2})}{2 b c}]$

= $\frac{1}{(k a b c)} . (b^{2} - c^{2}) (b^{2} + c^{2} - a^{2}) . . . . . (i)$

similarly, $\frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B = \frac{1}{(k a b c)} . (c^{2} - a^{2}) (c^{2} + a^{2} - b^{2}) . . . . . . (i i)$

And, $\frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = \frac{1}{(k a b c)} . (a^{2} - b^{2}) (a^{2} + b^{2} - c^{2}) .$

From (i), (ii) and (iii), we get

LHS = $\frac{b^{2} - c^{2}}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C$

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

= $\frac{1}{(k a b c)} \times 0 = 0 = R H S$

Hence, $\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = 0$

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Similar questions

Q. In any triangle

A B C

, prove that

\frac{b^{2} - c^{2}}{a^{2}} sin 2 A + \frac{c^{2} - a^{2}}{b^{2}} sin 2 B + \frac{a^{2} - b^{2}}{c^{2}} sin 2 C = 0

Q. Prove

\frac{b^{2} - c^{2}}{a^{2}} sin 2 A + \frac{c^{2} - a^{2}}{b^{2}} sin 2 B + \frac{a^{2} - b^{2}}{c^{2}} sin 2 C = 0

\frac{b^{2} - c^{2}}{a^{2}} s i n 2 A + \frac{c^{2} - a^{2}}{b^{2}} s i n 2 B + \frac{a^{2} - b^{2}}{c^{2}} s i n 2 C = 0

Q. Whether the given equation

(\frac{b^{2} - c^{2}}{a^{2}}) s i n 2 A + (\frac{c^{2} - a^{2}}{b^{2}}) s i n 2 B + (\frac{a^{2} - b^{2}}{c^{2}}) s i n 2 C = 1

?(In any

Δ

ABC )

In any $Δ A B C,$ prove that

$(b^{2} - c^{2}) c o t A + (c^{2} - a^{2}) c o t B + (A^{2} - B^{2}) c o t C = 0$

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In any Δ ABC, prove that (b2−c2)a2 sin 2A+(c2−a2)b2 sin 2B+(a2−b2)c2 sin 2C=0

In any $Δ A B C$ , prove that

$\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = 0$