Question

Prove that: 1/tan3A - tanA - 1/cot3A - cotA =cot2A

Answer 1

1/tan3A-tanA-1/cot3A-cotA
=1/{(3tanA-tan³A/1-3tan²A)-tanA}-1/{(cot³A-3cotA/3cot²A-1)-cotA}
=1/{(3tanA-tan³A-tanA+3tan³A)/(1-3tan²A)}-1/{(cot³A-3cotA-3cot³A+cotA)/(3cot²A-1)}
=(1-3tan²A)/2tanA(1+tan²A)-(3cot²A-1)/{-2(cotA+cot³A)}
=(1-3tan²A)/2tanA(1+tan²A)+(3/tan²A)-1}/{2(1/tanA+1/tan³A}
=(1-3tan²A)/2tanA(1+tan²A)+[{(3-tan²A)/tan²A}/{2(tan²A+1)/tan³A}]
=(1-3tan²A)/2tanA(1+tan²A)+tanA(3-tan²A)/2(1+tan²A)
={1-3tan²A+tan²A(3-tan²A)}/{2tanA(1+tan²A)}
=(1-3tan²A+3tan²A-tan⁴A)/2tanA(1+tan²A)
=(1-tan⁴A)/2tanA(1+tan²A)
={(1+tan²A)(1-tan²A)}/{2tanA(1+tan²A)}
=(1-tan²A)/2tanA
=cot2A (Proved)

Answer 2

Step-by-step explanation:

$\frac{1}{tan3a - tana} - \frac{1}{cot3a - cota} \\ \\ = \frac{1}{ \frac{sin3acosa - sinacos3a}{cos3acosa} } - \frac{1}{ \frac{cos3asina - cosasin3a}{sin3asina} } \\ \\ = \frac{cos3acosa}{sin(3a - a)} - \frac{sin3asina}{sin(a - 3a)} \\ \\ = \frac{cos3acosa}{sin2a} + \frac{sin3asina}{sin2a} \\ \\ = \frac{cos3acosa + sin3asina}{sin2a} \\ \\ = \frac{cos2a}{sin2a} \\ \\ = cot2a$