Inverse is defined for every trigonometric function in its domain.

True

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False

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Solution

The correct option is B False
We have seen while studying Relations and Functions that inverse is defined only for those functions which are bijective i.e. which are one - one and onto.
We also know that none of the trigonometric function is one - one in its domain.
We know right? Or Don’t we?
Well, we know all the trigonometric functions are periodic so more than one values of “x” give the same value of “y”. That’s why none of these functions are one - one. So inverse can’t be defined for these functions in their domains.
Inverse can be defined for these functions if we restrict their domains so that they remain as one - one function. For example Inverse can be defined for f(x) = sin(x) for f $[- \frac{π}{2}, \frac{π}{2}] \to [- 1, 1]$

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Similar questions

\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0

If a function is defined such that its inverse exists then the domain of the function becomes the range of the inverse function

\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

s i n^{- 1} x = | x - a |

What is inverse trigonometric functions?

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