Question

Choose the correct value for $x$ from the given options below if $x$ satisfies the inverse trigonometric equation $\sin \left( {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right) = \cos \left( {{{\tan }^{ - 1}}x} \right)$.

1. $\dfrac{1}{2}$
2. $1$
3. $0$
4. $\dfrac{{ - 1}}{2}$

Answer 1

Hint : We have given an Inverse trigonometric equation to solve. First, we take the Inverse trigonometric expressions and name them accordingly. And then we find the respective trigonometric ratios and equate them to solve. Finally, we end up with an algebraic equation to solve. We have to solve the algebraic equation to find the value of $x$.

Complete step-by-step answer :
Given an Inverse trigonometric equation to solve, that is
$\sin \left( {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right) = \cos \left( {{{\tan }^{ - 1}}x} \right)$,
Let us name the inverse trigonometric ratios in the equation first.
${\cot ^{ - 1}}\left( {1 + x} \right) = \theta$ , ${\tan ^{ - 1}}x = \varphi$
$\Rightarrow \cot \theta = 1 + x,\tan \varphi = x$
Now let us convert $\cot$ratio into $\tan$.
$\Rightarrow \tan \theta = \dfrac{1}{{1 + x}},\tan \varphi = x$ ,
Now let us convert both $\tan$ ratios into $\sin ,\cos$ ratios respectively.
$\Rightarrow \sin \theta = \dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }},\cos \varphi = \dfrac{1}{{\sqrt {1 + {x^2}} }}$ ,
Now let us apply ${\sin ^{ - 1}},{\cos ^{ - 1}}$ to the above equations respectively.
$\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }}} \right),\varphi = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)$ ,
Now let us replace the names with the original names that are,
$\Rightarrow {\cot ^{ - 1}}\left( {1 + x} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }}} \right),{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)$ ,
Now let us replace the above values in the given inverse trigonometric equation. We get,
$\Rightarrow \sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }}} \right)} \right) = \cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right)$ ,
We know that the function $\sin ,{\sin ^{ - 1}}$and $\cos ,{\cos ^{ - 1}}$ are pairs of inverse functions. So we get,
$\Rightarrow \dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }} = \dfrac{1}{{\sqrt {1 + {x^2}} }}$
So finally, we arrived at an algebraic equation.
Now to find the value of $x$ we need to solve this algebraic equation.
Let us reciprocal the given equation, we get
$\Rightarrow \sqrt {1 + {{\left( {1 + x} \right)}^2}} = \sqrt {1 + {x^2}}$,
Now let us square on both sides of the equation, we get
$\Rightarrow 1 + {\left( {1 + x} \right)^2} = 1 + {x^2}$,
Let us subtract $1$ on both sides, we get
$\Rightarrow {\left( {1 + x} \right)^2} = {x^2}$
Now expand,
$\Rightarrow 1 + {x^2} + 2x = {x^2}$
$\Rightarrow 1 + 2x = 0$
On further simplification, we get
$\Rightarrow x = \dfrac{{ - 1}}{2}$
Therefore the correct option is 4.
So, the correct answer is “Option 4”.

Note : This is a nice problem. At first, when we see the problem we try to perform some inverse trigonometric functions on the equation and find it clueless. That is the wrong method of solving this problem. This one way of solving this problem. The other way is to solve by using only trigonometric ratios, after we name the inverse trigonometric ratios we can find a relation from the given equation also, by using that equation we can solve the equations directly and end up with the same answer.