Question

Find the value of \[\sec \left( {{{\tan }^{ - 1}}\dfrac{y}{2}} \right)\] in terms of \[y\].

Answer 1

Hint: Here, we will first take \[\theta = {\tan ^{ - 1}}\dfrac{y}{2}\] in the given equation and then use the property of trigonometric functions, \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \] in the obtained equation. Then we will simplify the obtained expression to find the required value.

Complete step by step solution: We are given that the equation is \[\sec \left( {{{\tan }^{ - 1}}\dfrac{y}{2}} \right)\].

Let us take \[\theta = {\tan ^{ - 1}}\dfrac{y}{2}\] in the above equation.

\[ \Rightarrow \sec \theta {\text{ ......eq.(1)}}\]

We know that the property of trigonometric functions, \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \].

First, we will rewrite the above property of the trigonometric function by taking the square root on both sides.

\[ \Rightarrow \sec \theta = \sqrt {1 + {{\tan }^2}\theta } \]

Substituting this value of \[\sec \theta \] in the equation \[(1)\], we get

\[ \Rightarrow \sqrt {1 + {{\tan }^2}\theta } \]

Putting the value of \[\theta = {\tan ^{ - 1}}\dfrac{y}{2}\] back in the above equation, we get

\[ \Rightarrow \sqrt {1 + {{\left( {\tan \left( {{{\tan }^{ - 1}}\dfrac{y}{2}} \right)} \right)}^2}} \]

Using the property of the tangential property, \[{\tan ^{ - 1}}\tan x = x\] \[\forall x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\]in the above equation to find the value in terms of \[y\], we get

\[
   \Rightarrow \sqrt {1 + {{\left( {\dfrac{y}{2}} \right)}^2}} \\
   \Rightarrow \sqrt {1 + \dfrac{{{y^2}}}{4}} \\
   \Rightarrow \dfrac{{\sqrt {1 + {y^2}} }}{2} \\
 \]

Therefore, the value of \[\sec \left( {{{\tan }^{ - 1}}\dfrac{y}{2}} \right)\] in terms of \[y\] is \[\dfrac{{\sqrt {1 + {y^2}} }}{2}\].

Note: Here inverse trigonometric function was given in the question. many people misunderstand or don’t understand this function properly. It’s an angle so it should be treated as same. Also, the value inside it will be from its range. for example $ \theta = {\sin}^{-1} {x}$. Here $x$ is the reals value coming from the range of sin function.