Question

tan^-1 (x+1) + tan^-1 (x-1)= tan^1 (8/31)

Answer 1

Let 12arccos(ab)=y

Therefore, the required equation becomes

1+tan(y)1−tan(y)+1−tan(y)1+tan(y)

Simplifying the above expression, we get,

(1+tan(y))2+(1−tan(y))21−tan2(y)

=2sec2(y)1−tan2(y)

Dividing both numerator and denominator by sec2(y) we get,

2cos2(y)−sin2(y)

=2cos(2y)

Now,

cos(2y)=cos(arccos(ab))

cos(2y)=ab

Inputting the value of cos(2y) in our obtained expression, we get the required answer.

Hence, the required equation holds true.

Answer 2

Let 12arccos(ab)=y

Therefore, the required equation becomes

1+tan(y)1−tan(y)+1−tan(y)1+tan(y)

Simplifying the above expression, we get,

(1+tan(y))2+(1−tan(y))21−tan2(y)

=2sec2(y)1−tan2(y)

Dividing both numerator and denominator by sec2(y) we get,

2cos2(y)−sin2(y)

=2cos(2y)

Now,

cos(2y)=cos(arccos(ab))

cos(2y)=ab

Inputting the value of cos(2y) in our obtained expression, we get the required answer.

Hence, the required equation holds true.