(cosA+cosB)²+(sinA+sinB)²=4cos²(A-B /2)​

Solution:
LHS = (cosA+cosB)²+(sinA+sinB)²
= (cos²A+2cosAcosB+cos²B)+
(sin²A+2sinAsinB+sin²B)
/* By algebraic identity:
(x+y)² = x²+2xy+y² */
Rearranging the terms, we get
= (cos²A+sin²A)+(cos²B+sin²B)+2(cosAcosB+sinAsinB)
= 1+1+2cos(A-B)

we know that,
i) cos²x + sin²x = 1
ii) cosxcosy+sinxsiny = sin(x-y)

= 2+2cos(A-B)
= 2[1+cos(A-B)]

=
=
=$RHS$
Therefore,
(cosA+cosB)²+(sinA+sinB)²
=
••••

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