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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 x.

Explanation

Solution

The given equation sin⁡[cot⁡−1(x+1)]=cos⁡(tan⁡−1x)\sin \left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right) is easily solvable for ‘x’, if we try to convert the given inverse trigonometric function which gets easily cancel out with their respective trigonometric function, giving us an equation containing ‘x’. The equation is then solved to get the required value of ‘x’.

Complete step by step solution
Given: sin⁡[cot⁡−1(x+1)]=cos⁡(tan⁡−1x)\sin \left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right)
First of all, considering the given equation
sin⁡[cot⁡−1(x+1)]=cos⁡(tan⁡−1x)→(1)\sin \left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right) \to (1)
Let us take [cot⁡−1(x+1)]=a\left[ {{{\cot }^{ - 1}}\left( {x + 1} \right)} \right] = a
Then, cot⁡a=x+1\cot a = x + 1
As we know, a simple trigonometric equation –
cosec2a=1+cot⁡2a{{\mathop{\rm cosec}\nolimits} ^2}a = 1 + {\cot ^2}a
Thus, putting the value of we get –
coseca=1+(1+x)2{\mathop{\rm cosec}\nolimits} a = \sqrt {1 + {{\left( {1 + x} \right)}^2}}
Solving the equation, we get,
coseca=x2+2x+2{\mathop{\rm cosec}\nolimits} a = \sqrt {{x^2} + 2x + 2}
Since,
sin⁡a=1coseca∴sin⁡a=1x2+2x+2 \sin a = \dfrac{1}{{{\mathop{\rm cosec}\nolimits} a}} \therefore \sin a = \dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}
Hence, a=sin⁡−1(1x2+2x+2)=cot⁡−1(x+1)→(2)a = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} \right) = {\cot ^{ - 1}}\left( {x + 1} \right) \to \left( 2 \right)
Similarly, Let tan⁡−1x=b ⇒x=tan⁡b{\tan ^{ - 1}}x = b\\\ \Rightarrow x = \tan b
Knowing the simple trigonometric equation, sec⁡2b=1+tan⁡2b{\sec ^2}b = 1 + {\tan ^2}b
We get, sec⁡b=1+x2\sec b = \sqrt {1 + {x^2}}
Since,
cos⁡b=1sec⁡b ∴cos⁡b=11+x2\cos b = \dfrac{1}{{\sec b}}\\\ \therefore \cos b = \dfrac{1}{{\sqrt {1 + {x^2}} }}
Hence, b=cos⁡−111+x2=tan⁡−1x→(3)b = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }} = {\tan ^{ - 1}}x \to \left( 3 \right)
Putting the value of the equations (2) and (3), in that of (1), we get –
sin⁡[sin⁡−1(1x2+2x+2)]=cos⁡[cos⁡−111+x2] ⇒1x2+2x+2=11+x2 \sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} \right)} \right] = \cos \left[ {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right]\\\ \Rightarrow \dfrac{1}{{\sqrt {{x^2} + 2x + 2} }} = \dfrac{1}{{\sqrt {1 + {x^2}} }}
[Since, sin⁡−1(sin⁡θ)=θ∈(−π2,π2){\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right), cos⁡−1(cos⁡θ)=θ∈(0,π){\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta \in \left( {0,\pi } \right)]
Squaring and reciprocating on both sides,
We get,
x2+2x+2=1+x2{x^2} + 2x + 2 = 1 + {x^2}
Solving the equation, we get –
x=−12x = \dfrac{{ - 1}}{2}

**Thus, the required value of x=−12x = \dfrac{{ - 1}}{2}.

Note: **
The students should remember the basic trigonometric formulas and relations as they are very crucial in solving these types of questions.