How to proceed??

Question

How to proceed??

anonymous · Answer

$$Tan^-1(cotx)-Tan^-1(\cot2x) =???$$

anonymous · Answer

@hartnn @Luigi0210 @AkashdeepDeb @robtobey

anonymous · Answer

@amistre64 @Fifciol

akashdeepdeb · Answer

Do you know any formula to calculate
$$tan^{-1} (A) - tan^{-1} (B)~~~?$$

akashdeepdeb · Answer

@cambrige  ?

akashdeepdeb · Answer

Well, if there is any.

You may take cot x = A
thus, tan x = 1/A

And $$cot 2x = \frac{(1-tan^2 x)}{2 tan x}$$
                 $$ =  \frac{1  -  \frac{1}{A^2}}{\frac{2}{A}}$$

Calculate that and then use $tan^{-1} M - tan^{-1} N$ if there is a formula for that.
I think there is.

hartnn · Answer

yes
$\huge \tan^{-1}A-\tan^{-1}B = tan^{-1}(\dfrac{A-B}{1+AB})$

miracrown · Answer

|dw:1399190455867:dw|

well let me give you this general definition
following your directions:
what I wrote is that the cotangent of x is equal to 1 divided by the tangent of x
it will be very useful in solving this problem
so in addition to what I wrote on the board
here is the situation (in triangle terms) the cotangent is defined as the ratio of the adjacent side to the opposite side - hence the angle that I defined there
theta as I defined the sides accordingly
now we have that in place and we are looking for the inverse tangent of the ratio of those sides namely we are looking for the ratio of the angle
where the angle is this: