Whats the value of this??
Trigonometric
options
A)pi/4
B)-pi/4
C)pi/2
D)-pi/2
tan^−1 (x/y)−tan^−1 (x−y/x+y) =

Question

anonymous · Answer

$$\tan^{-1} (x/y)- \tan^{-1} (x-y/x+y) $$

anonymous · Answer

x,y>0

anonymous · Answer

@abb0t @hartnn @mathaddict4471

anonymous · Answer

@Loser66 @CGGURUMANJUNATH

abb0t · Answer

y ≠ 0, x = 0

anonymous · Answer

options are $$\pi/4..............(-\pi/4) ..........\pi/2 .............(-\pi/2)$$

anonymous · Answer

@agent0smith @BulletWithButterflyWings @Hero @ladydeadpool @mathaddict4471 @phi @primeralph @Psymon @Preetha

anonymous · Answer

@amistre64

primeralph · Answer

You don't have an equality sign anywhere.

tkhunny · Answer

I presume you mean (x-y)/(x+y) and not what you have written?

anonymous · Answer

the above equals which option???

primeralph · Answer

@cambrige  You were asked a question by @tkhunny

anonymous · Answer

No

anonymous · Answer

the answer to @tkhunny is no :)

tkhunny · Answer

You mean $\left(x - \dfrac{y}{x} + y\right)$, exactly as you have it written?

tkhunny · Answer

Well, $atan\left(\dfrac{x}{y}\right)-atan\left(\dfrac{x-y}{x+y}\right) = \dfrac{\pi}{4}$, but you go ahead and work the other problem that has no such value.

tkhunny · Answer

It may be simpler to show this:

$\sin\left(atan\left(\dfrac{x}{y}\right)-atan\left(\dfrac{x-y}{x+y}\right)\right) = \dfrac{1}{\sqrt{2}}$

The result requested follows immediately.

anonymous · Answer

whats atan??

anonymous · Answer

it is how u have put it @tkhunny

tkhunny · Answer

How about that.  It happens.  :-)

$atan(x)$ is, for most purposes, exactly the same as $tan^{-1}(x)$.  It's just easier to write.  One day, you may be concerned about a fundamental difference between the two, but I'm guessing that day is not today.