Question

tan(55°-A) -cot(35°+A)

Answer 1

we have here theta(0-)=A
therefore
tanA =cot(90-A). (1)
therefore
tan(55-A)-cot(35+A)=tan55-tanA-cot(35+A)
=tan55-cot(90-A)-cot35+cotA....... from (1)
=cot(90-55)-cot(90-A)-cot35+cotA.....since tanA =cot(90-A)
=cot35-cot(90-A)-cot35+cotA
= cotA-cot(90-A)
=0