Question

prove that cos9x-cos7x/sin9x-sin7x = -tan8x

Answer 1

Answer:

Formulas :

1)sin(C+D)=2sin(

2

C+D

)cos(

2

C−D

)

2)cos(C+D)=2cos(

2

C+D

)cos(

2

C−D

)

Rewriting the above equation ;

(cos9x+cos3x)+(cos7x+cos5x)

(sin9x+sin3x)+(sin7x+sin5x)

⇒

2cos(6x)cos(2x)+2cos(6x)cos(x)

2sin(6x)cos(2x)+2sin(6x)cos(x)

⇒

cos6x

sin6x

=tan6x

Answer 2

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

Consider,

LHS

$\rm :\longmapsto\:\dfrac{cos9x - cos7x}{sin9x - sin7x}$

We know,

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

and

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

So,

On applying these Identities, we get

$\rm \: = \: \: \:\dfrac{ - 2 \: sin\bigg(\dfrac{9x + 7x}{2} \bigg) \: sin\bigg(\dfrac{9x - 7x}{2} \bigg) }{2 \: cos\bigg(\dfrac{9x + 7x}{2} \bigg) \: sin\bigg(\dfrac{9x - 7x}{2} \bigg) }$

$\rm \: = \: \: \:\dfrac{ - \: sin\bigg(\dfrac{16x}{2} \bigg) \: sin\bigg(\dfrac{2x}{2} \bigg) }{ \: cos\bigg(\dfrac{16x}{2} \bigg) \: sin\bigg(\dfrac{2x}{2} \bigg) }$

$\rm \: = \: \: \: - \: \dfrac{sin8x}{cos8x}$

$\rm \: = \: \: \: - \: tan8x$

$\bf :\longmapsto\:\dfrac{cos9x - cos7x}{sin9x - sin7x} = - \: tan8x$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional information :-

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

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

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

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

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