Let f:A→B be a function defined by y=f(x) where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping g:B→A such that f(x)=y if and only if g(y)=x∀xϵA,yϵB Then function g is said to be inverse of f and vice versa so we write g=f−1:B→A[{f(x),x}:{x,f(x)}ϵf−1]when branch of an inverse function is not given (define) then we consider its principal value branch.If −1

Question

Let f:A→B be a function defined by y=f(x) where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping g:B→A such that f(x)=y if and only if g(y)=x∀xϵA,yϵB Then function g is said to be inverse of f and vice versa so we write g=f−1:B→A[{f(x),x}:{x,f(x)}ϵf−1]when branch of an inverse function is not given (define) then we consider its principal value branch.If −1<x<0,then tan−1x equals?

Filo · Accepted Answer

Let tan−1x=α⇒4−π​<α<0 ∴tanα=x such that 4−π​<α<0 ∴sinα=1+x2​x​⇒α=sin−1(1+x2​x​) =cot−1( x1​) =sec−11+x2​= cosec−1x1+x2​​ α=tan−1x=sin−1(1+x2​x​)

Question Text	Let $f : A \to B$ be a function defined by $y = f (x)$ where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping $g : B \to A$ such that $f (x) = y$ if and only if $g (y) = x \forall x ϵ A, yϵ B$ Then function g is said to be inverse of f and vice versa so we write $g = f^{- 1} : B \to A [{f (x), x} : {x, f (x)} ϵ f^{- 1}]$ when branch of an inverse function is not given (define) then we consider its principal value branch. If $- 1 < x < 0$ ,then $tan^{- 1} x$ equals?
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