Question

How do you solve $\csc \left( {{{\tan }^{ - 1}}\left( { - 2} \right)} \right)$ ?

Answer 1

Hint : In the given problem, we are required to calculate cosecant of an angle whose tangent is given to us in the question itself. Such problems require basic knowledge of trigonometric ratios and formulae. Besides this, knowledge of concepts of inverse trigonometry is extremely essential to answer these questions correctly.

Complete step by step solution:
So, In the given problem, we have to find the value of $\csc \left( {{{\tan }^{ - 1}}\left( { - 2} \right)} \right)$ .
Hence, we have to find the cosecant of the angle whose tangent is given to us as $- 2$ .
Let us assume $\theta$ to be the concerned angle.
Then, $\theta = {\tan ^{ - 1}}\left( { - 2} \right)$
We know that the principal value branch of ${\tan ^{ - 1}}x$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ . Since the value of tangent is negative. Hence, the angle must be in the fourth quadrant.
Taking tangent on both sides of the equation, we get
$\Rightarrow \tan \theta = \left( { - 2} \right)$
Since we have deduced that the angle $\theta$ lies in the fourth quadrant. So, there is no need for the negative sign.
So, we will first find the magnitude of the cosecant function and then adjust the sign as required.
To evaluate the value of the required expression, we must keep in mind the formulae of basic trigonometric ratios.
We know that, $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$ and $\csc \theta = \dfrac{{Hypotenuse}}{{Perpendicular}}$ .
So, $\left| {\tan \theta } \right| = \dfrac{{Perpendicular}}{{Base}} = 2$
So, let the length of the perpendicular be $2x$ .
Then, length of Base $= x$ .
Now, applying Pythagoras Theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}$
$\Rightarrow {\left( {Hypotenuse} \right)^2} = {\left( x \right)^2} + {\left( {2x} \right)^2}$
$\Rightarrow {\left( {Hypotenuse} \right)^2} = {x^2} + 4{x^2}$
$\Rightarrow {\left( {Hypotenuse} \right)^2} = 5{x^2}$
$\Rightarrow Hypotenuse = \sqrt {5{x^2}} = \sqrt 5 x$
So, we get $Hypotenuse = \sqrt 5 x$ .
Hence, $\csc \theta = \dfrac{{\sqrt 5 x}}{{2x}}$
$\Rightarrow \csc \theta = \left( {\dfrac{{\sqrt 5 }}{2}} \right)$
But, we know that angle $\theta$ lies in the fourth quadrant. Also, the cosecant function is negative in the fourth quadrant. Hence, $\csc \theta = - \dfrac{{\sqrt 5 }}{2}$
So, the value of $\csc \left( {{{\tan }^{ - 1}}\left( { - 2} \right)} \right)$ is $- \dfrac{{\sqrt 5 }}{2}$ .
So, the correct answer is “ $- \dfrac{{\sqrt 5 }}{2}$ ”.

Note : The given problem can also be solved by use of some basic trigonometric identities such as $\cos e{c^2}\theta = {\cot ^2}\theta + 1$ and ${\sec ^2}\theta = {\tan ^2}\theta + 1$ . This method also provides exposure to the applications of trigonometric identities in various mathematical questions.