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

In Δ ABC, by sine rule, we have sin A/a=sin B/b=sin C/c=k Then, sin A = ak, sin B = bk and sin C = ck Now, LHS = (b^2 c^2)cot A+(c^2 a^2)cot B+(a^2 b^2)cot C = ...

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Byju's Answer

Standard XI

Mathematics

Inequality

In a Δ ABC, p...

Question

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

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Solution

In $Δ A B C,$ by sine rule, we have $\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = k$

Then, sin A = ak, sin B = bk and sin C = ck

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

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

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

$[∵ u s i n g c o s i n e r u l e, c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c} a n d c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}]$

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

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

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

Hence proved.

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Similar questions

In any $Δ A B C,$ prove that

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

(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

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)

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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In a ΔABC, prove that (b2−c2)cot a+(c2−a2)cot B+(a2−b2)cot C=0.

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