In ABC- b-c-a B - 2 - C - 2 -2a B - 2

The correct option is B False (b+c a)(B/2+C/2)=(a+b+c 2 a)(s(s b)Δ+s(s c)Δ) #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard IX

Mathematics

Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

In ABC, b+c-...

Question

$I n △ A B C, (b + c - a) (cot \frac{B}{2} + cot \frac{C}{2}) = 2 a cot \frac{B}{2}$

True

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False

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Solution

The correct option is B False
$(b + c - a) (cot \frac{B}{2} + cot \frac{C}{2}) = (a + b + c - 2 a) (\frac{s (s - b)}{Δ} + \frac{s (s - c)}{Δ})$
$= (2 s - 2 a) \times \frac{s}{Δ} \times (s - b + s - c)$
$= 2 \times \frac{s (s - a)}{Δ} \times (2 s - b - c)$
$= 2 \times \frac{s (s - a)}{Δ} \times (a + b + c - b - c)$
$= 2 a cot \frac{A}{2}$
So the given relation is $False$

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Similar questions

Prove that $∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} c a - b^{2} & a b - c^{2} & b c - a^{2} a b - c^{2} & b c - a^{2} & c a - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is divisible by (a+b+c) and find the quotient.

Q. If

A = (1, 2, 3), B = (2, 3, 4)

and

A B

is produced upto

C

such that

2 A B = B C

, then

C =

Q. Factorise:

a^{2} - 2 a b + b^{2} - c^{2}

Q. Evaluate :

a^{2} + 2 a b + b^{2} - c^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c)^{2} & a^{2} & a^{2} b^{2} & (c + a)^{2} & b^{2} c^{2} & c^{2} & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 a b (a + b + c)^{3}

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In△ABC, (b+c−a)(cotB2+cotC2)=2acotB2

The correct option is B False(b+c−a)(cotB2+cotC2)=(a+b+c−2a)(s(s−b)Δ+s(s−c)Δ) =(2s−2a)×sΔ×(s−b+s−c) =2×s(s−a)Δ×(2s−b−c) =2×s(s−a)Δ×(a+b+c−b−c) =2acotA2So the given relation is False

$I n △ A B C, (b + c - a) (cot \frac{B}{2} + cot \frac{C}{2}) = 2 a cot \frac{B}{2}$