The domain of trigonometric functions can be restricted to any one of their branches (not necessarily principal value) in order to obtain their inverse functions. State True/False

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Solution

Check the bijective nature of functions in different branches of domain

$∵$ Trigonometric functions are many-one in their domains.

So, in order to define the inverse trigonometric function, we need to restrict the domain in such a way that the function is one-one and onto.

$∵$ We know that trigonometric functions are one-one and onto in many of the branches in Its domain.

$∴$ We can restrict domain to any one of the branches.

Hence, the statement is true.

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All trigonometric functions have inverse over their respective domains.

State True/False.

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\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

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