In any - ABC- prove that b 2- c 2 A - c 2- a 2 B - A 2- B 2 C -0

By the sine rule, we have a/sin A = b/sin B = c/sin C ⇒sin A/a = sin B/b = sin C/c = k (say) ⇒ sin a = ka, sin b = kb and sin C = kc. ∴ (b^2 c^2) cot A = (b^2 ...

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Standard XII

Mathematics

Cosine Rule

In any Δ ABC,...

Question

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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Solution

By the sine rule, we have

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}$

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

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

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

= $\frac{1}{(2 b a b c)} . (b^{2} - c^{3}) (b^{2} + c^{2} - a^{2}) = \frac{b^{4} c^{4} - a^{2} b^{2} + a^{2} c^{2}}{2 k a b c}$

Similiary, $(c^{2} - a^{2}) c o t B = \frac{(c^{4} a^{4} - b^{2} c^{2} + a^{2} b^{2})}{2 k a b c} .$

And, $(a^{2} - b^{2}) c o t C = \frac{(a^{4} - b^{4} - a^{2} c^{2} + b^{2} c^{2})}{2 k a b c} .$

$∴ L H S = (b^{2} - c^{2}) c o t A + (c^{2} - a^{2}) c o t B + (a^{2} - b^{2}) c o t C$

= $\frac{1}{2 k a b c} . [(b^{4} - c^{4} - a^{2} b^{2} + a^{2} c^{2}) + (c^{4} - a^{2} - b^{2} c^{2} + a^{2} b^{2}) + (a^{4} - b^{4} - a^{2} c^{2} + b^{2} c^{2})]$

= $\frac{1}{2 k a b c} \times 0 = 0 = R H S$

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

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Similar questions

In a $Δ A B C,$ prove that $(b^{2} - c^{2}) c o t a + (c^{2} - a^{2}) c o t B + (a^{2} - b^{2}) c o t C = 0.$

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

Q. In

△ A B C, (b^{2} - c^{2}) cot A + (c^{2} - a^{2}) cot B + (a^{2} - b^{2}) cot C

In any $Δ A B C$ , prove that

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

Q. Factorize :

a^{2} (b + c)^{2} + b^{2} (c + a)^{2} + c^{2} (a + b) + a b c (a + b + c) + (a^{2} + b^{2} + c^{2}) (b c + c a + a b)

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In any ΔABC, prove that (b2−c2) cot A+(c2−a2) cot B+(A2−B2) cot C=0

In any $Δ A B C,$ prove that

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