Question

Trigonometry.

Answer 1

If
\[\huge 3\sin \alpha=5\sin \beta\]
. then
\[\huge \frac{ \tan \frac{ \alpha+\beta }{ 2 } }{ \tan \frac{ \alpha-\beta }{ 2 } }\] equals

Answer 2

Just need a hint

Answer 3

@SithsAndGiggles

Answer 4

give meh a sec

Answer 5

Ok

Answer 6

\[\frac{1}{\frac{2\sin \left(α\right)}{\sin \left(α\right)+\sin \left(β\right)}-1}\]

Answer 7

what is that expression

Answer 8

Hint :-
sin A + sin B =2 sin (A+B/2) cos (A-B /2)
sin A - sin B =2 sin (A-B/2) cos (A+B /2)
mmm (hope it work i still dint try it yet :P)

Answer 9

I know the formulae ,but how to use them here lol

Answer 10

@ganeshie8

Answer 11

Someone help me please

Answer 12

hehe ok xD
iactully im reading a book , but found this interesting
so ill start on my note :D

Answer 13

This is a ridiculous topic

Answer 14

sh*T

Answer 15

:o

Answer 16

Options:-
(A) 1 (B) 2 (C) 3 (D)4

Answer 17

ikram's formulas wil lwork

Answer 18

\[\large \frac{ \tan \frac{ \alpha+\beta }{ 2 } }{ \tan \frac{ \alpha-\beta }{ 2 } }\]


\[\large \frac{ \sin \frac{ \alpha+\beta }{ 2 } \cos\frac{ \alpha-\beta }{ 2 } }{ \cos\frac{ \alpha+\beta }{ 2 } \sin\frac{ \alpha-\beta }{ 2 } }\]

Answer 19

\[\large \frac{2 \sin \frac{ \alpha+\beta }{ 2 } \cos\frac{ \alpha-\beta }{ 2 } }{ 2\cos\frac{ \alpha+\beta }{ 2 } \sin\frac{ \alpha-\beta }{ 2 } }\]

Answer 20

fine, so far ?

Answer 21

go it after that

Answer 22

got*

Answer 23

use this formula :
2sinAcosB = sin(A+B) + sin(A-B)

Answer 24

thank you @ganeshie8 and @ikram002p ^_^

Answer 25

yeah

Answer 26

Oh you figured out the answer ?

Answer 27

yeah , kind off , the method i got it now

Answer 28

okie :)