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Question: Range of the function \[y = f\left( x \right) = \dfrac{{1 - tanx}}{{1 + tanx}}\] ​ is- A.R -{1} ...

Range of the function y=f(x)=1−tanx1+tanxy = f\left( x \right) = \dfrac{{1 - tanx}}{{1 + tanx}} ​ is-
A.R -{1}
B.R -{-1,1}
C.R
D.R -{-1}

Explanation

Solution

Hint : To get the range of the given trigonometric expression we will find the x in terms of y. then we apply the condition of existence of mathematical functions. Such that we will get the range of the given function.

Complete step-by-step answer :
Given that
f(x)=1−tanx1+tanxf\left( x \right) = \dfrac{{1 - tanx}}{{1 + tanx}}
suppose the f(x) as y
y=f(x)=1−tanx1+tanxy = f\left( x \right) = \dfrac{{1 - tanx}}{{1 + tanx}}
on cross multiplying the terms we get
y+y×tanx=1−tanxy + y\times tanx = 1 - tanx
taking y as common
tanx(y+1)=1−ytanx\left( {y + 1} \right) = 1 - y
bring all y terms to one side
tanx=1−y1+ytanx = \dfrac{{1 - y}}{{1 + y}}
finding x
f−1(x)=tan−1(1−y1+y){f^{ - 1}}(x) = ta{n^{ - 1}}\left( {\dfrac{{1 - y}}{{1 + y}}} \right)
as f−1(x){f^{ - 1}}\left( x \right) is same as x
so,
f−1(x)=tan−1(1−y1+y){f^{ - 1}}(x) = ta{n^{ - 1}}\left( {\dfrac{{1 - y}}{{1 + y}}} \right)
f−1(x){f^{ - 1}}\left( x \right) valid if   (1+x)≠0⇒x≠−1\;\left( {1 + x} \right) \ne 0 \Rightarrow x \ne - 1
We can say that f−1(x){f^{ - 1}}\left( x \right) is valid for all real number except -1
So the domain of f−1(x){f^{ - 1}}\left( x \right) is R - \left\\{ { - 1} \right\\}
Therefore the range of f(x)  f\left( x \right)\; is R - \left\\{ { - 1} \right\\}
So, the correct answer is “Option D”.

Note : There are many methods to find the range of a mathematical function. Such as graphical method and by knowing the domain and finding the corresponding range. Here in this solution we have used the inverse approach which is also an easy way to find the range of the given function.