Question

In a triangle, prove that (b+c-a)(cotB/2+cotC/2)=2a×cotA/2

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\displaystyle \sf{(b + c - a) \bigg( \: \cot \frac{B}{2} + \cot \frac{C}{2} \bigg) = 2a \cot \frac{A}{2} \: }$

PROOF

For a triangle a, b, c are the sides and

perimeter = 2s

$\sf{So \: \: \: 2s = a+b+c}$

$\sf{ \implies \: 2s - a= b+c}$

LHS

$= \displaystyle \sf{(b + c - a) \bigg( \: \cot \frac{B}{2} + \cot \frac{C}{2} \bigg) \: }$

$= \displaystyle \sf{(2s - a - a) \bigg[\: \frac{s(s - b)}{ \Delta} + \frac{s(s - c)}{ \Delta} \bigg ] \: }$

$= \displaystyle \sf{(2s - 2a ) \bigg[\: \frac{s(s - b)}{ \Delta} + \frac{s(s - c)}{ \Delta} \bigg ] \: }$

$= \displaystyle \sf{ \frac{2s( s - a)}{ \Delta} \bigg[\: s - b + s - c \bigg ] \: }$

$= \displaystyle \sf{ \frac{2s( s - a)}{ \Delta} \bigg[\: 2 s - b - c \bigg ] \: }$

$= \displaystyle \sf{ \frac{2s( s - a)}{ \Delta} \bigg[\: a + b + c - b - c \bigg ] \: }$

$= \displaystyle \sf{ \frac{2s( s - a)}{ \Delta} \: \times a\: }$

$= \displaystyle \sf{2a \times \frac{s( s - a)}{ \Delta} \: }$

$\displaystyle \sf{ = 2a \cot \frac{A}{2} \: }$

Hence proved

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LEARN MORE FROM BRAINLY

tan⁻¹2 +tan⁻¹3 is =

Select Proper option from the given options.

(a) - π/4

(b) π/2

(c) 3π/4

(d) 3π/2

https://brainly.in/question/5596487