Hola Brainlians. Here’s a question to check your Trigonometric skill. ICSE Board : Inverse Trigonometry, Class 10. Give answer fast and no spam.
Proof the given identity.
tan^{-1}x – tan^{-1} y = tan^{-1} [(x – y)/(1 + xy)] , xy > – 1
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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
Solution!!
To prove:-
[tex]\sf \tan ^{-1}x-\tan ^{-1}y=\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)[/tex]
First of all, we will
Let [tex]\sf \tan ^{-1}x\:be\:\alpha[/tex]
And
Let [tex]\sf \tan ^{-1}y\:be\:\beta[/tex]
We will get the values of x and y.
x = [tex]\sf \tan \alpha[/tex]
y = [tex]\sf \tan \beta[/tex]
Now let’s subtract!!
[tex]\sf \tan (\alpha -\beta )=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }[/tex]
[tex]\sf \alpha -\beta =\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)[/tex]
[tex]\sf \tan ^{-1}x -\tan ^{-1}y=\tan ^{-1}\left(\dfrac{x-y}{1+xy}\right)[/tex]
Hence, proved!