Question

show that sinA+sin5A+sin9A/cosA+cos5S+cos9A=tan5A

Answer 1

This is the answer... We have to substitute appropriate formula and thus we get final answer

Answer 2

Answer:

tan 5A

Explanation:

Here, the given equation is

$\dfrac{sin A + sin 5A + sin 9 A}{cos A + cos 5A + cos 9A}$

we know the formula of sin C + sin D and $sin C + sin D = sin \dfrac{C+D}{2} cos \dfrac{C-D}{2} \\and\\cos C + cos D = cos \dfrac{C+D}{2} cos \dfrac{C-D}{2}$

apply the formula in the given equation

$\dfrac{sin A + sin 5A + sin 9 A}{cos A + cos 5A + cos 9A}[tex]=\dfrac{sin A + sin 9A + sin 5A}{cos A + cos 9A + cos 5A}\\\\=\dfrac{sin\frac{(9A+A)}{2}cos\frac{(9A-A)}{2} + sin5A }{cos\frac{(9A+A)}{2}cos\frac{(9A-A)}{2} + cos5A } \\=\dfrac{sin5A cos 8A+sin 5A}{cos5Acos8A+cos5A} \\=\dfrac{sin5A(cos8A+1)}{cos5A(cos8A+1)} \\=\dfrac{sin5A}{cos5A} \\=tan5A$

Hence proved.