Question

If 3 sinA=5 sinB, then find the value of —
(tan (A+B)/2)/( tan( A-B)/2)​

Answer 1

Answer:

4

Step-by-step explanation:

3sinA = 5sinB   =>  $\frac{sinA}{sinB}$ = $\frac{5}{3}$

tan$(\frac{A+B}{2} )$  / tan $(\frac{A-B}{2})$

can be written as [sin$(\frac{A+B}{2} )$ / cos$(\frac{A+B}{2} )$ ] * [cos$(\frac{A-B}{2} )$ / sin$(\frac{A-B}{2} )$ ]  

=  sin$(\frac{A+B}{2} )$ cos$(\frac{A-B}{2} )$ /cos$(\frac{A+B}{2} )$ sin$(\frac{A-B}{2} )$

= $\frac{2(sinA + sinB)}{2(sinA - sinB)}$

Divide by sinB

i.e  ( $\frac{sinA}{sinB} + 1$ ) / ( $\frac{sinA}{sinB} -1$ )

= ( $\frac{5}{3} -1$ ) / ( $\frac{5}{3} -1$ )

= ($\frac{5+3}{3}$) /  ($\frac{5-3}{3}$)

=$\frac{8}{2}$ =  4