Question

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

Hint: In this question, we will proceed by converting ${\cot ^{ - 1}}$ in terms of ${\sin ^{ - 1}}$ and ${\tan ^{ - 1}}$ in terms of ${\cos ^{ - 1}}$ by using the formula ${\cot ^{ - 1}}\left( {x + 1} \right) = {\sin ^{ - 1}}\dfrac{1}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }}$ and ${\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}$. So, use this concept to reach the solution of the problem.

Given that $\sin \left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right)$
By using the formula ${\cot ^{ - 1}}\left( {x + 1} \right) = {\sin ^{ - 1}}\dfrac{1}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }}$ and ${\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}$
$\Rightarrow \sin \left[ {{{\sin }^{ - 1}}\dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }}} \right] = \cos \left[ {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right]$
We know that, $\sin \left( {{{\sin }^{ - 1}}A} \right) = A$ and $\cos \left( {{{\cos }^{ - 1}}A} \right) = A$
$\Rightarrow \dfrac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }} = \dfrac{1}{{\sqrt {1 + {x^2}} }} \\ \Rightarrow \sqrt {1 + {x^2}} = \sqrt {1 + {{\left( {1 + x} \right)}^2}} \\$
$\Rightarrow 1 + {x^2} = 1 + {\left( {1 + x} \right)^2} \\ \Rightarrow 1 + {x^2} = 1 + 1 + 2x + {x^2} \\ \Rightarrow 1 + {x^2} = 2 + 2x + {x^2} \\ \Rightarrow 1 + {x^2} - 2 - {x^2} = 2x \\ \Rightarrow - 1 = 2x \\ \therefore x = - \dfrac{1}{2} \\$
Thus, the value of $x$ is $- \dfrac{1}{2}$
Note: To solve these kinds of questions, students must be familiar with all the formulae in trigonometry and inverse trigonometry. If we didn’t remember the formulae we can draw the corresponding right angle triangle then we convert the terms of ${\cot ^{ - 1}}$ and ${\tan ^{ - 1}}$ in terms of ${\sin ^{ - 1}}$ and ${\cos ^{ - 1}}$ respectively. But it consumes a lot of time. So, do remember the formulae in order to solve them easily.