Question

Find the solution set of the equation below, the equation is:
 \[ta{{n}^{-1}}x-co{{t}^{-1}}x=co{{s}^{-1}}\left( 2-x \right)\].

Answer 1

Hint: Take cosine at both the sides of the equation. Then apply the concept that \[\cos \left( s-t \right)=\cos \left( s \right)\cos \left( t \right)+\sin \left( s \right)\sin \left( t \right)\] and also that
  \[\begin{align}
  & \Rightarrow ta{{n}^{-1}}x=\theta \\
 & \Rightarrow \tan \theta =x \\
\end{align}\]
Also, we have:
\[\Rightarrow \cos \left( \theta \right)=\dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\]
\[\Rightarrow \sin \,\,\theta =\dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\]

Complete step by step answer:
In the question, we have to solve the given equation \[ta{{n}^{-1}}x-co{{t}^{-1}}x=co{{s}^{-1}}\left( 2-x \right)\].
So, here we will start by taking cosine both the sides of the above equation, to obtain:
\[
 \Rightarrow \cos \left( ta{{n}^{-1}}x-co{{t}^{-1}}x \right)=\cos \left( {{\cos }^{-1}}\left( 2-x \right) \right) \\
 \Rightarrow \cos \left( ta{{n}^{-1}}x \right)\cos \left( co{{t}^{-1}}x \right)+\sin \left( ta{{n}^{-1}}x \right)\sin \left( co{{t}^{-1}}x \right)=\cos \left( {{\cos }^{-1}}\left( 2-x \right) \right) \\
\]
Because we have: \[\cos \left( s-t \right)=\cos \left( s \right)\cos \left( t \right)+\sin \left( s \right)\sin \left( t \right)\]. Next, we will find the value of
Left hand side \[\cos \left( ta{{n}^{-1}}x \right)\cos \left( co{{t}^{-1}}x \right)+\sin \left( ta{{n}^{-1}}x \right)\sin \left( co{{t}^{-1}}x \right)\]
Now, since we have
\[
   \Rightarrow ta{{n}^{-1}}x=\theta \\
  \Rightarrow \tan \theta =x \\
\]
So here, we have
\[
   \Rightarrow \cos \left( ta{{n}^{-1}}\left( x \right) \right)=\dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}} \\
  \Rightarrow \cos \left( {{\cot }^{-1}}\left( x \right) \right)=\dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}} \\
  \Rightarrow \sin \left( ta{{n}^{-1}}\left( x \right) \right)=\dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}} \\
  \Rightarrow \sin \left( {{\cot }^{-1}}\left( x \right) \right)=\dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}} \\
\]
Next, the expression at the left-hand side and right-hand side is solved as shown below:
\[\dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\cdot \dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}+\dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\cdot \dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}=2-x\]
Since we have \[\cos \left( {{\cos }^{-1}}\left( 2-x \right) \right)=2-x\]
Now, we will solve it as follows:
\[
 \Rightarrow \dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\cdot \dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}+\dfrac{x\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\cdot \dfrac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}=2-x \\
  \Rightarrow 2{{\left( \sqrt{1+{{x}^{2}}} \right)}^{2}}x=2{{\left( 1+{{x}^{2}} \right)}^{2}}-x{{\left( 1+{{x}^{2}} \right)}^{2}} \\
  \Rightarrow 2x+2{{x}^{3}}=2+4{{x}^{2}}+2{{x}^{4}}-x-2{{x}^{3}}-{{x}^{5}} \\
  \Rightarrow -{{x}^{5}}+2{{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}-3x+2=0 \\
  \Rightarrow -\left( x-1 \right)\left( {{x}^{4}}-{{x}^{3}}+3{{x}^{2}}-x+2 \right)=0 \\
\]
Now here we see that the expression \[\left( {{x}^{4}}-{{x}^{3}}+3{{x}^{2}}-x+2 \right)=0\] has no solution and the second equation is
 \[
   \Rightarrow \left( x-1 \right)=0 \\
  \Rightarrow x=1 \\
\]

So here we will have only one solution and that is at \[x=1\]

Note: Care has to be taken when applying the formula for the expansion form of \[\cos \left( s-t \right)=\cos \left( s \right)\cos \left( t \right)+\sin \left( s \right)\sin \left( t \right)\], we have to take care of the positive and the negative sign. So, this is a bit different from the expansion of \[\sin \left( s-t \right)\]. We have to solve the equation using the squaring of the terms of the equation.