Given sin α + sin β =A And cos α + cos β =B.

⇒A² +B² =(sin α + sin β)² +(cos α + cos β)²

= sin² α + sin² β + 2 sin α sin β + cos² α + cos² β + 2 cos α cos β

= sin² α + cos² α + sin² β + cos² β + 2(sin α sin β + cos α cos β)

We know that cos(A -B) = cosA cosB + sinA sinB

∴A² +B² = 2 + 2 cos(α – β) …(1)

Then,

⇒B² –A² =(cos α + cos β)² –(sin α + sin β)²

= cos² α + cos² β + 2 cos α cos β –(sin² α + sin² β + 2 sin α sin β)

=(cos² α – sin² β) +(cos² β – sin² α) – 2cos(α + β)

= 2 cos(α + β) cos(α – β) + 2 cos(α + β)

= cos(α + β)(2 + 2 cos(α – β)) …(2)

From(1) And(2),

⇒B² –A² = cos(α + β)(A² +B²)

…(ii)

And

…(i)