Question

How do you evaluate \[\sec \left( {{{\cot }^{ - 1}}\left( { - 2} \right)} \right)\] without a calculator?

Answer 1

Hint: To solve this question first we assume some angle equal to the angle given in the function and then we convert inverse trigonometry function to simple trigonometry function. Then we put the value of sec function in terms of tan function and then on further calculation we get the final value of the given expression.

Complete answer:
To find,
The value of the given expression \[\sec \left( {{{\cot }^{ - 1}}\left( { - 2} \right)} \right)\].
Let the value of the given expression is-
\[x = \sec \left( {{{\cot }^{ - 1}}\left( { - 2} \right)} \right)\] …….(i)
Let \[x = \sec \theta \] ……(ii)
On comparing equation (i) and (ii) the value of \[\theta \] is
\[\theta = {\cot ^{ - 1}}\left( { - 2} \right)\]
Now taking cot function both side
\[\cot \theta = \cot ({\cot ^{ - 1}}( - 2))\]
We know that \[\cot ({\cot ^{ - 1}}\theta ) = \theta \]
On putting this equation in the obtained equation
\[\cot \theta = - 2\]
Now on converting it into tan function we know that \[\tan \theta = \dfrac{1}{{\cot \theta }}\]
\[\tan \theta = - \dfrac{1}{2}\]
Now in order to find the value of \[x\] we have to find the value of \[\sec \theta \].
So now we have to express sec function in terms of tan function.
\[x = \sec \theta \]
Now using the identity of trigonometry \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
\[x = \sqrt {1 + {{\tan }^2}\theta } \]
Now putting the value of the \[\tan \theta \]
\[x = \sqrt {1 + {{\left( { - \dfrac{1}{2}} \right)}^2}} \]
Onn further calculations
\[x = \sqrt {1 + \dfrac{1}{4}} \]
On taking LCM
\[x = \sqrt {\dfrac{{4 + 1}}{4}} \]
\[x = \sqrt {\dfrac{5}{4}} \]
On simplifying
\[x = \dfrac{{\sqrt 5 }}{2}\]
Final answer:
The value of the given expression is \[\sec \left( {{{\cot }^{ - 1}}\left( { - 2} \right)} \right)\] is-
\[ \Rightarrow x = \dfrac{{\sqrt 5 }}{2}\]

Note:
To solve these types of questions we must know all the identities and the formula of trigonometry and inverse trigonometry. Always try to take a variable as an angle to the inverse part. Students commit mistakes in making inverse trigonometry to simple trigonometry functions. And they are unable to use their identity. We also find the value of sec function directly from the value of the tan function by using the Pythagoras theorem on a right angle triangle.