Question & Answer

If \[\sin \left[ {{\cot }^{-1}}(x+1) \right]=\cos \left( {{\tan }^{-1}}x \right)\] then find x.

ANSWER Verified Verified
Hint: Put \[{{\cot }^{-1}}\left( x+1 \right)=a\] and \[{{\tan }^{-1}}x=b\]. Substitute these values in their respective trigonometric formula connecting \[\left( \cos ec\theta ,\cot \theta \right)\] and \[\left( sec\theta ,tan\theta \right)\]. Thus get the value for a and b. Substitute and solve for x.

Complete step-by-step answer:

Given to us is the expression \[\sin \left[ {{\cot }^{-1}}\left( x+1 \right) \right]=\cos \left(

{{\tan }^{-1}}x \right)......(1)\]

Let us consider the LHS of the equation.

Let us put \[{{\cot }^{-1}}\left( x+1 \right)=a\]


  & \therefore {{\cot }^{-1}}\left( x+1 \right)=a \\

 & x+1=\cot a......(2) \\


We know the trigonometric formula,


  & \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1 \\

 & \therefore \cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \\

 & \therefore \cos ec\theta =\sqrt{1+{{\cot }^{2}}\theta } \\


Thus, \[\cos eca=\sqrt{1+{{\cot }^{2}}a}\]

Put cot a = x + 1 in the above equation.

\[\cos eca=\sqrt{1+{{\left( x+1 \right)}^{2}}}\]

We know \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]

\[\therefore \cos eca=\sqrt{1+{{x}^{2}}+2x+1}=\sqrt{{{x}^{2}}+2x+2}\]

We know \[\sin a={}^{1}/{}_{\cos eca}\].


  & \therefore \sin a=\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}} \\

 & \therefore a={{\sin }^{-1}}\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}.......(3) \\


Now let us consider the RHS of the equation.

Put \[{{\tan }^{-1}}x=b\].

\[\therefore x=\tan b\]

We know that \[{{\sec }^{2}}b-{{\tan }^{2}}b=1\]


  & \therefore {{\sec }^{2}}b=1+{{\tan }^{2}}b \\

 & \sec b=\sqrt{1+{{\tan }^{2}}b} \\


Put \[\tan b=x\].

\[\therefore \sec b=\sqrt{1+{{x}^{2}}}\]

We know \[\cos b={}^{1}/{}_{\sec b}\].


  & \therefore \cos b=\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\

 & \therefore b={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right).......(4) \\


We have to put \[{{\cot }^{-1}}(x+1)=a\] and \[{{\tan }^{-1}}x=b\]. Thus equation (1) becomes,


  & \sin \left[ {{\cot }^{-1}}(x+1) \right]=\cos \left( {{\tan }^{-1}}x \right) \\

 & \Rightarrow \sin a=\cos b \\


Put the value of equation (3) and (4) in the above expression.

\[\sin \left[ {{\sin }^{-1}}\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}} \right]=\cos \left[ {{\cos

}^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}} \right]\]

Cancel out sin and \[{{\sin }^{-1}}\] from LHS and cos and \[{{\cos }^{-1}}\] from RHS.

We get,


  & \dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}=\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\

 & \sqrt{1+{{x}^{2}}}=\sqrt{{{x}^{2}}+2x+2} \\

 & {{\left( \sqrt{1+{{x}^{2}}} \right)}^{2}}={{\left( \sqrt{{{x}^{2}}+2x+2} \right)}^{2}} \\

 & \Rightarrow 1+{{x}^{2}}={{x}^{2}}+2x+2 \\


Cancel out common terms.


  & 2x+2=1 \\

 & 2x=1-2 \\

 & x={}^{-1}/{}_{2}. \\


Thus we got the value of \[x={}^{-1}/{}_{2}.\]

Note: You can also do it using the trigonometric identity \[{{\cot }^{-1}}\theta ={{\sin }^{-1}}\dfrac{1}{\sqrt{1+{{\theta }^{2}}}}\].
Put \[\theta =x+1\].
\[\therefore {{\cot }^{-1}}(x+1)={{\sin }^{-1}}\dfrac{1}{\sqrt{1+{{\left( x+1 \right)}^{2}}}}\]
Similarly, \[{{\tan }^{-1}}\theta ={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{A}^{2}}}}\]. Put \[\theta =x\].
\[\therefore {{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}\]
Thus substitute the value of \[{{\cot }^{-1}}(x+1)\] and \[{{\tan }^{-1}}x\] and you get value of x as \[{}^{-1}/{}_{2}\].