Question

cot+cos=m and cot - cos=n prove that (m^2-n^2)^2= 16

Answer 1

2cot=m+n
2cos=m-n
l.h.s=[m²-n²]²
[(m+n)(m-n)]²
[2cot.2cos]²
[4cot.cos]²
16(cot²).(cos²)
I think the question is wrong...
it should be
(m²-n²)²=16mn
so now,
r.h.s=16mn
=16(cot+cos)(cot-cos)
=16(cot²-cos²)
=16{(cos²/sin²)-cos²}
=16cos²{(1/sin²)-1}
=16cos²{(1-sin²)/sin²}
=16cos²{cos²/sin²}
=16cos².cot²
therefore,
l.h.s=r.h.s