is sinA =sinB & cosA =cosB,then
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sinA−sinB=0 and cosA−cosB=0
⇒ 2sin
2
A−B
cos
2
A+B
=0
and 2sin
2
A+B
cos
2
B−A
=0
We observe that the common factor gives sin
2
A−B
=0.
Thus,
2
A−B
=nπ, n∈Z
A−B=2nπ, n∈Z
A=2nπ+B, n∈Z
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