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# If $cosec\left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{7}} \right) + {{\cot }^{ - 1}}\left( x \right)} \right] = 1$, find the value of $x$.

Last updated date: 22nd Jul 2024
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Hint: Here, in the question, we have been given a trigonometric equation including an unknown variable and we are asked to find the value of that variable. On the right hand side of the equation, $1$ is given and on the left hand side, function of cosecant is given. We will put the cosecant function on the right hand such that the angle of cosecant gives the value $1$. And then, we will simplify using inverse trigonometric identity and get the desired result.
Formula used:
${\tan ^{ - 1}}\dfrac{1}{x} = {\cot ^{ - 1}}x \\ {\cot ^{ - 1}}\left( { - x} \right) = \pi - {\cot ^{ - 1}}x \\ \cot \left( {{{\cot }^{ - 1}}x} \right) = x \\ \cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot y + \cot x}}$

Complete step-by-step solution:
Given equation: $cosec\left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{7}} \right) + {{\cot }^{ - 1}}\left( x \right)} \right] = 1$
Now, we have a cosecant function on the left hand side. To make the equation simpler, we will have to bring the cosecant function on the right hand side. For that, we will put the cosecant function such that the angle of the cosecant results in the value $1$. And we know, at $\dfrac{\pi }{2}$ cosecant function is equal to $1$.
Therefore, the equation becomes as,
$cosec\left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{7}} \right) + {{\cot }^{ - 1}}\left( x \right)} \right] = cosec\left( {\dfrac{\pi }{2}} \right) \\ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} \\$
Using the identity, ${\tan ^{ - 1}}\dfrac{1}{x} = {\cot ^{ - 1}}x$, we get,
${\cot ^{ - 1}}\left( 7 \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} \\ \Rightarrow {\cot ^{ - 1}}\left( 7 \right) - \dfrac{\pi }{2} = - {\cot ^{ - 1}}\left( x \right)$
Using identity, $\left[\because {\cot ^{ - 1}}\left( { - x} \right) = \pi - {\cot ^{ - 1}}x \Rightarrow - {\cot ^{ - 1}}\left( x \right) = {\cot ^{ - 1}}\left( { - x} \right) - \pi \right]$ ,
we get,
${\cot ^{ - 1}}\left( 7 \right) - \dfrac{\pi }{2} = {\cot ^{ - 1}}\left( { - x} \right) - \pi \\$
Adding $\pi$ both sides, we get,
${\cot ^{ - 1}}\left( 7 \right) + \dfrac{\pi }{2} = {\cot ^{ - 1}}\left( { - x} \right)$
Taking $\cot$ both sides, we get,
$\cot \left( {{{\cot }^{ - 1}}\left( 7 \right) + \dfrac{\pi }{2}} \right) = \cot \left( {{{\cot }^{ - 1}}\left( { - x} \right)} \right) \\ \Rightarrow \cot \left( {{{\cot }^{ - 1}}\left( 7 \right) + \dfrac{\pi }{2}} \right) = - x \\$ $\left[ {\because \cot \left( {{{\cot }^{ - 1}}x} \right) = x} \right]$
Using identity $\cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot y + \cot x}}$, we obtain,
$\dfrac{{\cot \left( {{{\cot }^{ - 1}}7} \right)\cot \left( {\dfrac{\pi }{2}} \right) - 1}}{{\cot \left( {\dfrac{\pi }{2}} \right) + \cot \left( {{{\cot }^{ - 1}}7} \right)}} = - x$
Now, we know that, $\cot \dfrac{\pi }{2} = 0$, then,
$\dfrac{{ - 1}}{7} = - x$
Multiplying by $\left( { - 1} \right)$ both sides, we get,
$x = \dfrac{1}{7}$
Hence, the value of $x$ is $\dfrac{1}{7}$.

Note: The symbol ${\tan ^{ - 1}}x$ should not be confused with ${\left( {\tan x} \right)^{ - 1}}$. In-fact ${\tan ^{ - 1}}x$ is an angle, the value of whose tangent is $x$. The only key concept to solve such types of questions is that we must remember all the basic trigonometric identities and their application. This will help us to solve almost all the questions.