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,
sin2A−cos2B=sin2(

π

3

)−sin2(

π

6

)
=

3

4

−

1

4

=

1

2

Therefore, sin2A−sin2B=

1

2