UPSC NDA Math Model Paper 4

Given that sin(A+B)=1 and sin(A−B)=

1

2

.
In the given domain the function sin‌x is one one.
Therefore, A+B=

π

2

and A−B=

π

6

.
Solving the both equations for A we get,
2A=

4π

6

A=

π

3

Therefore, A=

π

3

.
Substituting in the value of A+ B we get that B=

π

6

Therefore,
tan(A+2B)⋅tan(2A+B) =tan(

π

3

+2(

π

6

)).tan(

π

6

+2(

π

3

))
=tan(

2π

3

).tan(

5π

6

)
=tan(

π

2

+

π

6

).tan(

π

2

+

π

3

)
=−cot(

π

6

).−cot(

π

3

)
=√3.

1

√3

=1
Therefore, tan(A+2B)⋅tan(2A+B)=1