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Question: The number of solutions for the equation \[{{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\le...

The number of solutions for the equation tan1(ex)+cot1(lnx)=π2{{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2} is:
(a) 0
(b) 1
(c) 3
(d) 2

Explanation

Solution

We solve this problem first by taking the trigonometric equations out by using the standard formulas that is
tan1x=cot1(1x){{\tan }^{-1}}x={{\cot }^{-1}}\left( \dfrac{1}{x} \right)
Then we use the composite angle formula of inverse trigonometric equations as
cot1x+cot1y=cot1(xy1x+y){{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)
By using the above formulas we find the relation between to functions without trigonometric equations so that we can find the number of solutions using the graphs that is if f(x)=g(x)f\left( x \right)=g\left( x \right) then the number of solutions to above equations will be number of points of intersections of y=f(x)y=f\left( x \right) and y=g(x)y=g\left( x \right)

Complete step-by-step solution
We are given that the equation that is
tan1(ex)+cot1(lnx)=π2\Rightarrow {{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2}
We know that the inverse trigonometric relation that is
tan1x=cot1(1x){{\tan }^{-1}}x={{\cot }^{-1}}\left( \dfrac{1}{x} \right)
By using this relation to above equation we get

& \Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{{{e}^{-x}}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2} \\\ & \Rightarrow {{\cot }^{-1}}\left( {{e}^{x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2} \\\ \end{aligned}$$ We know that the formula of the composite angle formula of inverse trigonometric equations as $${{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)$$ By using this formula to above equation we get $$\begin{aligned} & \Rightarrow {{\cot }^{-1}}\left( \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|} \right)=\dfrac{\pi }{2} \\\ & \Rightarrow \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|}=\cot \dfrac{\pi }{2} \\\ \end{aligned}$$ We know that from the standard table of trigonometric ratios that is $$\cot \dfrac{\pi }{2}=0$$ By substituting this value in above equation we get $$\begin{aligned} & \Rightarrow \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|}=0 \\\ & \Rightarrow {{e}^{x}}\left| \ln x \right|-1=0 \\\ & \Rightarrow \left| \ln x \right|={{e}^{-x}} \\\ \end{aligned}$$ We know that if $$f\left( x \right)=g\left( x \right)$$ then the number of solutions to above equations will be number of points of intersections of $$y=f\left( x \right)$$ and $$y=g\left( x \right)$$ Now, let us draw the graph of $$y=\left| \ln x \right|$$ and $$y={{e}^{-x}}$$ then we get Here we can see that the blue line represents $$y={{e}^{-x}}$$ and the green line represents $$y=\left| \ln x \right|$$ Here, we can see that both the graphs intersect at 2 points therefore the number of solutions of the given equation is 2. So, option (d) is the correct answer. ![](https://www.vedantu.com/question-sets/4f6c4525-3f19-4f67-aab3-b013a7726ef17369997381406994470.png) **Note:** We have a shortcut for solving the problem. We are given that the equation as $$\Rightarrow {{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2}$$ We have a standard result of inverse trigonometric ratios that is if $${{\tan }^{-1}}x+{{\cot }^{-1}}y=\dfrac{\pi }{2}$$ happens if and only if $$x=y$$ By using the above result we get $$\Rightarrow {{e}^{-x}}=\left| \ln x \right|$$