# Hard Inverse Trig/Induction Question (1 Viewer)

#### Riviet

##### .
Prove by mathematical induction for all positives,
tan-1[1/(2 x 12)] + tan-1[1/(2 x 22)] + ... + tan-1[1/(2n2)] = pi/4 - tan-1[1/(2n+1)]

I have proven it for n=1, but stuck with the proving for n=k+1 part.
Any help would be greatly appreciated, thanks. Last edited:

#### tutor01

##### Member
Here is the n=k+1 part of the solution.

See attached.

#### Riviet

##### .
Ah so that's how you do it, thanks a bunch for the fully worked solution, really apprecate it! #### .ben

##### Member
I don't get the bit where you put let alpha, and then tan(alpha)?
also how would you rearrange this expression: tan(tan-1a-tan-1b)

can you pllease explain it thanks

#### Mountain.Dew

##### Magician, and Lawyer.
.ben said:
I don't get the bit where you put let alpha, and then tan(alpha)?
also how would you rearrange this expression: tan(tan-1a-tan-1b)

can you pllease explain it thanks
all tutor1 has done is simply made that big tan inverse expression as one angle. remember that trig inverse (something) is merely the value of an angle, nothing else.

he uses tan(alpha) so that he can get rid of the tan inverses and work with the actual algebraic expressions inside the trig inverse.

so, we have...tan(tan-1a-tan-1b)

suppose we let tan-1a = A and tan-1b = B

SO we know that tanA = tan(tan-1a) = a and tanB = tan(tan-1b) = b

SO, tan(tan-1a-tan-1b) = tan(A-B) = tanA - tanB / (1-tanAtanB) = (a - b) / (1 - ab)

#### yungaten

##### New Member
Here is the n=k+1 part of the solution.

See attached.
Where is the attached file?