Question

the value of cos12A - cos 2A / sin 12A + sin 2A​

Answer 1

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

$\rm :\longmapsto\:\dfrac{cos12A - cos2A}{sin12A + sin2A}$

We know,

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

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

Using these two Identities, we get

$\: \: \rm= \: \:\dfrac{ - 2sin\bigg(\dfrac{12A + 2A}{2} \bigg)sin\bigg(\dfrac{12A - 2A}{2} \bigg)}{2sin\bigg(\dfrac{12A + 2A}{2} \bigg)cos\bigg(\dfrac{12A - 2A}{2} \bigg)}$

$\: \: \rm= \: \:\dfrac{ - sin\bigg(\dfrac{14A}{2} \bigg)sin\bigg(\dfrac{10A}{2} \bigg)}{sin\bigg(\dfrac{14A}{2} \bigg)cos\bigg(\dfrac{10A}{2} \bigg)}$

$\: \: \rm= \: \:\dfrac{ - \cancel{\: sin7A} \: sin5A}{ \cancel{sin7A} \: \: cos5A}$

$\: \: \rm= \: \: - \: \dfrac{sin5A}{cos5A}$

$\: \: \rm= \: \: - \: tan5A$

Hence,

$\bf :\longmapsto\:\dfrac{cos12A - cos2A}{sin12A + sin2A} = - \: tan5A$

Additional Information :-

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

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

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

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

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