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

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

Explanation

Solution

In this question, we will proceed by converting cot⁡−1{\cot ^{ - 1}} in terms of sin⁡−1{\sin ^{ - 1}} and tan⁡−1{\tan ^{ - 1}} in terms of cos⁡−1{\cos ^{ - 1}} by using the formula cot⁡−1(x+1)=sin⁡−111+(x+1)2{\cot ^{ - 1}}\left( {x + 1} \right) = {\sin ^{ - 1}}\dfrac{1}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }} and tan⁡−1x=cos⁡−111+x2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}. So, use this concept to reach the solution of the problem.

Complete step by step answer:
Given that sin⁡[cot⁡−1(x+1)]=cos⁡(tan⁡−1x)\sin \left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right)
By using the formula cot⁡−1(x+1)=sin⁡−111+(x+1)2{\cot ^{ - 1}}\left( {x + 1} \right) = {\sin ^{ - 1}}\dfrac{1}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }} and tan⁡−1x=cos⁡−111+x2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}
⇒sin⁡[sin⁡−111+(1+x)2]=cos⁡[cos⁡−111+x2]\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⁡(sin⁡−1A)=A\sin \left( {{{\sin }^{ - 1}}A} \right) = A and cos⁡(cos⁡−1A)=A\cos \left( {{{\cos }^{ - 1}}A} \right) = A

⇒11+(1+x)2=11+x2 ⇒1+x2=1+(1+x)2  \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}} \\\

Squaring on both sides, we have

⇒1+x2=1+(1+x)2 ⇒1+x2=1+1+2x+x2 ⇒1+x2=2+2x+x2 ⇒1+x2−2−x2=2x ⇒−1=2x ∴x=−12  \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 xx is −12 - \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{\cot ^{ - 1}} and tan⁡−1{\tan ^{ - 1}} in terms of sin⁡−1{\sin ^{ - 1}} and cos⁡−1{\cos ^{ - 1}} respectively. But it consumes a lot of time. So, do remember the formulae in order to solve them easily.