Question

{1/tan 3A+tanA }-{1/cot3A+cot A}​

Answer 1

Let us write t for tan A.

It may help to know that

tan 3A=(3t-t^3)/(1–3t^2)

tan 3A - tan A = (2t+2t^3)/(1–3t^2)

cot 3A = (1–3t^2)/(3t-t^3)

cot 3A-cot A=(1–3t^2)/(3t-t^3)-1/t

=(-2–2t^2)/(3t-t^3)

1/(tan3A-tan A)-1/(cot 3A - cot A)

=(1–3t^2)/(2t(1+t^2))

+(3t-t^3)/(2(1+t^2))

=(1-t^4)/(2t(1+t^2))

=(1-t^2)/(2t)

=cot 2A.

Answer 2

$\frac{1}{tan3a + tana} - \frac{1}{cot3a + cota} \\ \\ = \frac{1}{ \frac{sin3acosa + sinacos3a}{cos3acosa} } - \frac{1}{ \frac{cos3asina + cosasin3a}{sin3asina} } \\ \\ = \frac{cos3acosa}{sin(3a + a)} - \frac{sin3asina}{sin(a + 3a)} \\ \\ = \frac{cos3acosa - sin3asina}{sin4a} \\ \\ = \frac{cos4a}{sin4a} \\ \\ = cot4a$