Question

sin A + sin 5A + sin 9A ÷ cos A + vos 5A + cos 9A = tan 5A

Answer 1

Answer:

Tan5A

Step-by-step explanation:

L.H.S

cosA+cos5A+cos9AsinA+sin5A+sin9A

We know that

sinC+sinD=2sin(2C+D)⋅cos(2C−D)

cosC+cosD=2cos(2C+D)⋅cos(2C−D)

 

Therefore,

=2cos(2A+9A)⋅cos(2A−9A)+cos5A2sin(2A+9A)⋅cos(2A−9A)+sin5A

=2cos(5A)⋅cos(4A)+cos5A2sin(5A)⋅cos(4A)+sin5A

=2cos(4A)+12cos(4A)+1(cos5Asin5A

Answer 2

Answer:

Here given for prove,

$\frac{ \sin(A) + \sin(5A) + \sin(9A) }{ \cos(A) + \cos(5A) + \cos(9A) } = \tan(A)$

Here,

L.H.S$= \frac{ \sin(A) + \sin(5A) + \sin(9A) }{ \cos(A) + \cos(5A) + \cos(9A) }$

and R.H.S$= \tan(A)$

We know,

$\sin(C) + \sin(D) = 2 \sin( \frac{C+ D}{2} ) \cos( \frac{C - D}{2} )$

and

$\cos(C) + \cos(D) = 2 \cos( \frac{C + D}{2} ) \cos( \frac{C - D}{2} )$

Therefore,LH.S$= \frac{ \sin(A) + \sin(5A) + \sin(9A) }{ \cos(A) + \cos(5A) + \cos(9A) }$

$= \frac{2 \sin( \frac{A + 9A}{2} ) \cos( \frac{A - 9A}{2} ) +sin( 5A) }{2 \cos( \frac{A + 9A}{2} ) \cos( \frac{A - 9A}{2} ) + cos(5A) }$

$= \frac{2 \sin(5A) . \cos(4A) + \sin(5A) }{2 \cos(5A) . \cos(4A) + \cos(5A)}$

$= \frac{2 \cos(4A) + 1 }{2 \cos(4A) + 1} \times \frac{ \sin(5A) }{ \cos(5A) }$

$= \frac{ \sin(5A) }{ \cos(5A) }$

$= \tan(5A)$

= R.H.S

Therefore L.H.S=R.H.S [proved]