Question

How do you find the value of \[\sec \left[ {{{\cot }^{ - 1}}\left( { - 6} \right)} \right]\]?

Answer 1

Hint: In this question we have to find the value of the given function which is a trigonometric function, we will make use of inverse trigonometric formulas, first convert cot function in terms of tan using identity \[{\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)\], and then convert tan function in terms of cos by using identity \[{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)\], then using trigonometric identity \[{\cos ^{ - 1}}x = {\sec ^{ - 1}}\dfrac{1}{x}\], then simplify the expression using the formula \[{\sec ^{ - 1}}\left( {\sec x} \right) = x\], we will get the required result.

Complete answer:
Given function is \[\sec \left[ {{{\cot }^{ - 1}}\left( { - 6} \right)} \right]\],
First using the fact \[{\cot ^{ - 1}}\left( { - x} \right) = \pi - {\cot ^{ - 1}}x\], the expression becomes,
\[ \Rightarrow \sec \left[ {\pi - {{\cot }^{ - 1}}6} \right]\],
Now using the identity \[\sec \left( {\pi - x} \right) = - \sec x\], we get,
\[ \Rightarrow - \sec \left( {{{\cot }^{ - 1}}6} \right)\],
Now convert cot function in terms of tan using the identity \[{\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)\], we get,
Here \[x = 6\], now substituting the value in the formula we get,
\[ \Rightarrow - \sec \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{6}} \right)} \right)\],
Now convert tan function in terms of cos by using identity \[{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)\], we get,
Here \[x = \dfrac{1}{6}\], now substituting the value in the formula we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {{\left( {\dfrac{1}{6}} \right)}^2}} }}} \right)} \right)\],
Now simplifying we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + \dfrac{1}{{36}}} }}} \right)} \right)\],
Now taking L.C.M in the denominator we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{{\sqrt {\dfrac{{36 + 1}}{{36}}} }}} \right)} \right)\],
Now simplifying we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{{\sqrt {\dfrac{{37}}{{36}}} }}} \right)} \right)\],
Now taking the denominator of the denominator to the numerator we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{{\sqrt {36} }}{{\sqrt {37} }}} \right)} \right)\],
Now simplifying we get,
\[ \Rightarrow - \sec \left( {{{\cos }^{ - 1}}\left( {\dfrac{6}{{\sqrt {37} }}} \right)} \right)\],
Now using the identity, \[{\cos ^{ - 1}}x = {\sec ^{ - 1}}\dfrac{1}{x}\], here \[x = \dfrac{6}{{\sqrt {37} }}\], substituting the value in the formula we get,
\[ \Rightarrow - \sec \left( {{{\sec }^{ - 1}}\left( {\dfrac{{\sqrt {37} }}{6}} \right)} \right)\],
Now we know that \[ \Rightarrow \sec \left( {{{\sec }^{ - 1}}\left( x \right)} \right) = x\], we get,
\[ \Rightarrow - \sec \left( {{{\sec }^{ - 1}}\left( {\dfrac{{\sqrt {37} }}{6}} \right)} \right) = - \dfrac{{\sqrt {37} }}{6}\],
So, the value of the given function is \[ - \dfrac{{\sqrt {37} }}{6}\].

Final Answer:
\[\therefore \]The value of the given function \[\sec \left[ {{{\cot }^{ - 1}}\left( { - 6} \right)} \right]\]will be equal to \[ - \dfrac{{\sqrt {37} }}{6}\].

Note:
The inverse trigonometric functions are used to find the unknown measure of an angle of a right triangle when two side lengths are known. Basic inverse trigonometric functions are \[{\sin ^{ - 1}}x\], \[{\cos ^{ - 1}}x\] and \[{\tan ^{ - 1}}x\].Specifically, they are the inverses of the sine, cosine, tangent. Some important formulas related to Inverse trigonometry are given below:
\[{\sin ^{ - 1}}\left( {\sin x} \right) = x\],
\[{\cos ^{ - 1}}\left( {\cos x} \right) = x\]
\[{\tan ^{ - 1}}\left( {\tan x} \right) = x\],
\[{\sec ^{ - 1}}\left( {\sec x} \right) = x\],
\[{\csc ^{ - 1}}\left( {\csc x} \right) = x\],
\[{\cot ^{ - 1}}\left( {\cot x} \right) = x\],
\[{\sin ^{ - 1}}x = {\csc ^{ - 1}}\dfrac{1}{x}\],
\[{\cos ^{ - 1}}x = {\sec ^{ - 1}}\dfrac{1}{x}\],
\[{\tan ^{ - 1}}x = {\cot ^{ - 1}}\dfrac{1}{x}\],
\[{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right)\],
\[{\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right)\]
\[{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)\].