Question

please answer q25 with detailed solution

Answer 1

$\large\underline{\sf{Given \:Question - }}$

If

$\rm :\longmapsto\:A + B + C = \pi$

Prove that

$\rm :\longmapsto\:\dfrac{cosA}{sinBsinC} + \dfrac{cosB}{sinCsinA} + \dfrac{cosC}{sinAsinB} = 2$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:\dfrac{cosA}{sinBsinC} + \dfrac{cosB}{sinCsinA} + \dfrac{cosC}{sinAsinB}$

$\rm \: = \:\dfrac{sinAcosA + sinBcosB + sinCcosC}{sinAsinBsinC}$

On multiply and divide by 2, we get

$\rm \: = \:\dfrac{2sinAcosA +2 sinBcosB +2 sinCcosC}{2sinAsinBsinC}$

We know,

$\boxed{ \bf{ \: sin2x = 2sinxcosx}}$

So, using this we get

$\red{\rm \: = \:\dfrac{sin2A + sin2B + sin2C}{2sinAsinBsinC} - - - (1)}$

Now, Consider,

$\rm :\longmapsto\:sin2A + sin2B + sin2C$

We know,

$\boxed{ \bf{ \: sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$

So, on using this identity, we get

$\rm \: = \:2sin\bigg(\dfrac{2A + 2B}{2} \bigg)cos\bigg(\dfrac{2A - 2B}{2} \bigg) + sin2C$

$\rm \: = \:2sin(A +B)cos(A - B) + sin2C$

It is given that

$\rm :\longmapsto\:A + B + C = \pi$

So,

$\rm \: = \:2sin(\pi - C)cos(A - B) + sin2C$

$\rm \: = \:2sinCcos(A - B) + 2sinCcosC$

$\rm \: = \:2sinC\bigg(cos(A - B) + cosC\bigg)$

$\rm \: = \:2sinC\bigg(cos(A - B) + cos(\pi - (A + B))\bigg)$

$\rm \: = \:2sinC\bigg(cos(A - B) - cos(A + B)\bigg)$

We know,

$\boxed{ \bf{ \: cos(x - y) - cos(x + y) = 2sinx \: siny}}$

So, using this, we get

$\rm \: = \:2sinA(2sinBsinC)$

$\rm \: = \:4sinAsinBsinC$

So, we have

$\red{ \boxed{ \sf{ \:sin2A + sin2B + sin2C = 4sinAsinBsinC }}}$

So, from equation (1), we have

$\rm \: = \:\dfrac{sin2A + sin2B + sin2C}{2sinAsinBsinC}$

$\rm \: = \:\dfrac{4sinAsinBsinC}{2sinAsinBsinC}$

$\rm \: = \:2$

Hence,

$\rm :\longmapsto\: \boxed{ \bf{ \: \dfrac{cosA}{sinBsinC} + \dfrac{cosB}{sinCsinA} + \dfrac{cosC}{sinAsinB} = 2}}$

Additional Information :-

$\boxed{ \bf{ \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}}$

$\boxed{ \bf{ \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$

$\boxed{ \bf{ \: cosx - cosy = 2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{y - x}{2} \bigg)}}$

$\boxed{ \bf{ \: 2sinxcosy = sin(x + y) + sin(x - y)}}$

$\boxed{ \bf{ \: 2cosxcosy = cos(x + y) + cos(x - y)}}$