Question

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Proof the given identity.

tan^{-1}x - tan^{-1} y = tan^{-1} [(x - y)/(1 + xy)] , xy > - 1

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Answer 1

Solution!!

To prove:-

$\sf \tan ^{-1}x-\tan ^{-1}y=\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)$

First of all, we will

Let $\sf \tan ^{-1}x\:be\:\alpha$

And

Let $\sf \tan ^{-1}y\:be\:\beta$

We will get the values of x and y.

x = $\sf \tan \alpha$

y = $\sf \tan \beta$

Now let's subtract!!

$\sf \tan (\alpha -\beta )=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$

$\sf \alpha -\beta =\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)$

$\sf \tan ^{-1}x -\tan ^{-1}y=\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)$

Hence, proved!

Answer 2

Answer:

→ Let α=tan

−1

x,β=tan

−1

y

As tan(α+β)=

1−tanαtanβ

tanα+tanβ

⇒tan(α+β)=

1−xy

x+y

⇒α+β=tan

−1

(

1−xy

x+y

)= tan

−1

x+tan

−1

y