Question

sin A+sin 3 A+sin 5A/cos A+ cos 3A+ cos 5A= tan 3A​​

Answer 1

Step-by-step explanation:

$\huge\bf\red{\underline{\underline{ANSWER:-}}}$

From L . H . S -

$\sf{\dfrac{Sin A + Sin 3A + Sin 5A}{Cos A + Cos 3A + Cos 5A}}$

$\sf{=\dfrac{(Sin 5A + Sin A) + Sin 3A}{(Cos 5A + Cos A) + Cos 3A}}$

$\sf{=\dfrac{2Sin(\dfrac{5A + A}{2})Cos(\dfrac{5A - A}{2}) + Sin 3A}{2Cos(\dfrac{5A + A} {2})Cos(\dfrac{5A - A}{2}) + Cos 3A}}$

$\sf{=\dfrac{2Sin 3ACos 2A + Sin 3A}{2Cos 3ACos 2A + Cos 3A}}$

$\sf{=\dfrac{Sin 3A(2Cos 2A + 1)}{Cos 3A(2Cos 2A + 1)}}$

$\sf{=\dfrac{Sin 3A}{Cos 3A}}$

$\sf{= tan 3A}$

$\sf{= R . H . S}$

★ Using Formulas :-

$\sf{SinC + SinD = 2Sin(\dfrac{C + D}{2})Cos(\dfrac{C - D}{2})}$

$\sf{CosC + CosD = 2Cos(\dfrac{C + D}{2})Cos(\dfrac{C - D}{2})}$