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How do you find the inverse of arc $\tan (x + 3)$ and is it a function?

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f^{- 1} (x) = tanx + 3

f^{- 1} (x) = - 3 + tanx

f^{- 1} (x) = tanx - 3

f^{- 1} (x) = - tanx - 3

answer is C.

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Detailed Solution

We know that arc tan represents the inverse trigonometric function

ta n^{- 1}

, so let's assume the given expression as y, we have,
⇒

y = ta n^{- 1} (x + 3)

As the range of

ta n^{- 1}

function is (

\frac{- π}{2}, \frac{π}{2}

), so here

y \in (\frac{- π}{2}, \frac{π}{2})

.
Taking tangent function both the sides, we get,
⇒

tan y = \tan [ta n^{- 1} (x + 3)]

⇒

tan y = x + 3

[\tan (ta n^{- 1} x) = x

]
⇒

x + 3 = tan y

⇒

x = tan y - 3

………… 1
Also,

ta n^{- 1} (x + 3)

is a function of

x

, therefore we can write it as

f (x)

, so, we get,
⇒

f (x) = ta n^{- 1} (x + 3)

⇒

f (x) = y

⇒

x = f^{- 1} (y)

………….2
From equations (1) and (2),
⇒

f^{- 1} (y) = tan y - 3

Replacing

y

with

x

,
⇒

f^{- 1} (x) = tan y - 3

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