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```
Sin inverse X + Cot inverse X + tan inverse X Range=????
Sin inverse X + Cot inverse X + tan inverse X Range=????

```
9 months ago

```							we know the standard identity:Cot inverse x + tan inverse x= pi/2also, Sin inverse x lies in [– pi/2, pi/2].so that range of Sin inverse X + Cot inverse X + tan inverse X= Sin inverse X + pi/2 is [0, pi].KINDLY APPROVE :))
```
9 months ago
```							f(x)=sin−1x+cos−1x+tan−1xUsing the Inverse Trigonometric Identity,sin−1x+cos−1x=π/2We get,f(x)=π/2+tan−1xNow,, range of tan−1x as [−π/2,π/2]…-π/2 so,-π/2+π/2 here it comes , 0 but here it comes a great mistake becauseNow most of us apply the range of tan−1x as [−π/2,π/2]So we get the range of f(x) as [0,π] but there is a very big mistake here.We know that the domain of f(x)f(x) is [−1,1] hence the term tan−1x wouldn’t hold the values −π/2 and π/2So the new range becomes,f(−1)=π/2−π/4=π/4 f(−1)=π2−π/4=π/4f(1)=π2+π/4=3π/4 f(1)=π/2+π/4=3π/4So we get f(x)∈[π/4,3π/4]
```
9 months ago
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