If $\sin (A+B)=1$ and $\cos (A-B)=\frac{\sqrt{3}}{2}, A+B \leq 90^{\circ}$ and $A>B$, then the value of $\frac{5 \sin ^2 B+4 \tan ^2 A}{2 \sin B \cos A}$ is :

$26 \frac{1}{2}$

We are given :-

sin ( A + B ) = 1 

{ we know, sin 90º = 1 }

So, ( A + B ) = 90º    ----(1)

& cos ( A - B ) = \(\frac{√3}{2}\)

{ cos 30º = \(\frac{√3}{2}\) }

So,  ( A - B ) = 30º     ----(2)

On adding equation 1 and 2 .

2A = 120º

A = 60º

Put value of A in equation 1 .

60º + B = 90º 

B = 30º

Now,

\(\frac{5sin²B + 4tan²A }{2sinB.cosA}\)

= \(\frac{5sin²30º + 4tan²60º }{2sin30º .cos60º }\)

= \(\frac{5 (1/4) + 4 (3) }{2(1/2) . (1/2)  }\)

= \(\frac{53}{2  }\)

= 26\(\frac{1}{2  }\)