Question

prove that cotA.cot(60-A).cot(60+A)=cot3A

urgent i gave the brainlist answer​

Answer 1

cota+cot(60°+a)-cot(60°-a)

=cota+{(cot60°cota-1)/(cota+cot60°)}-{(cot60°cota+1)/(cota-cot60°)}

[∵, cot(A+B)=(cotBcotA-1)/(cotB+cotA) and cot(A-B)=(cotBcotA+1)/(cotB-cotA)]

=cota+[{(cota/√3)-1}/{cota+(1/√3)}]-[{(cota/√3)+1}/{cota-(1/√3)}]

=cota+{(cota-√3)/(√3cota+1)}-{(cota+√3)/(√3cota-1)}

=cota+{(cota-√3)(√3cota-1)-(cota+√3)(√3cota+1)}/{(√3cot+1)(√3cota-1)}

=cota+(√3cot²a-3cota-cota+√3-√3cot²a-3cota-cota-√3)/{(√3cota)²-(1)²}

=cota+(-8cota)/(3cot²a-1)

=(3cot³a-cota-8cota)/(3cot²a-1)

=3(cot³a-3cota)/(3cot²a-1)

=3[{(1/tan³a)-(3/tana)}/{(3/tan²a)-1}]

=3[{(1-3tan²a)/tan³a}/{(3-tan²a)/tan²a}]

=3{(1-3tan²a)/tana(3-tan²a)}

=3/{(3tana-tan³a)/(1-3tan²a)}

=3/tan3a  [∵, tan3a=(3tana-tan³a)/(1-3tan²a)]

=3cot3a(Proved)

Answer 2

Step-by-step explanation:

1/tanAtan(60-A)tan(60+A)

1/tan3A = Cot 3A

Hope this helps you