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

tan 2 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cot 3 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan 3 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

tan 4 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $tan 3 A$
$\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A}$
$= \frac{(sin A + sin 5 A) + (sin 4 A + sin 2 A)}{(cos 5 A + cos A) + (cos 4 A + cos 2 A)}$
$= \frac{2 sin 3 A cos 2 A + 2 sin 3 A cos A}{2 cos 3 A cos 2 A + 2 cos 3 A cos A}$
$= \frac{2 sin 3 A (cos 2 A + cos A)}{2 cos 3 A (cos 2 A + cos A)}$
$= tan 3 A$

Suggest Corrections

Similar questions

\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A} =

{cos}^{4} A - {cos}^{2} A = {sin}^{4} A - {sin}^{2} A

c o s^{4} A - c o s^{2} A = s i n^{4} A - s i n^{2} A

\frac{{cos}^{4} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

\frac{{cos}^{4} B}{{cos}^{2} A} + \frac{{sin}^{4} B}{{sin}^{2} A} = 1

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program