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

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

& \Rightarrow ta{{n}^{-1}}x=\theta \\\ & \Rightarrow \tan \theta =x \\\ \end{aligned}$$ Also, we have: $$\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.