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 :
Answer & explanation
Correct answer: option 4
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\(\frac{1}{2 }\)