QUESTION

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

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.

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

\begin{align} & \therefore {{\cot }^{-1}}\left( x+1 \right)=a \\ & x+1=\cot a......(2) \\ \end{align}

We know the trigonometric formula,

\begin{align} & \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 } \\ \end{align}

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 $\sin a={}^{1}/{}_{\cos eca}$.

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

Now let us consider the RHS of the equation.

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

$\therefore x=\tan b$

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

\begin{align} & \therefore {{\sec }^{2}}b=1+{{\tan }^{2}}b \\ & \sec b=\sqrt{1+{{\tan }^{2}}b} \\ \end{align}

Put $\tan b=x$.

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

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

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

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

\begin{align} & \sin \left[ {{\cot }^{-1}}(x+1) \right]=\cos \left( {{\tan }^{-1}}x \right) \\ & \Rightarrow \sin a=\cos b \\ \end{align}

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,

\begin{align} & \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 \\ \end{align}

Cancel out common terms.

\begin{align} & 2x+2=1 \\ & 2x=1-2 \\ & x={}^{-1}/{}_{2}. \\ \end{align}

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 ${}^{-1}/{}_{2}$.