Question

How do you simplify the expression $\sec \left( {{\tan }^{-1}}\left( x \right) \right)$

Answer 1

To simplify the expression we will first assume ${{\tan }^{-1}}x=y$ . Now we will apply tan function on both sides and simplify. Now we will try to convert tan into sec by using identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$ . Hence we will easily find the value of ${{\sec }^{2}}y$ and take square root to find the value of $\sec y$ . Now re-substituting the value of y we will get the required equation.

Complete step-by-step solution:
Now first let us understand the concept of inverse function. Inverse functions are functions which reverse the role of the original function. Hence if we have $f\left( x \right)=y$ then ${{f}^{-1}}\left( y \right)=x$ . Now from the two equations we can substitute the value of x to write $f\left( {{f}^{-1}}y \right)=y$ .
For example consider $f\left( x \right)=2x$ then ${{f}^{-1}}\left( x \right)=\dfrac{x}{2}$ .
Now let us consider $y={{\tan }^{-1}}x$
Then applying tan function on both sides we get, $\tan y=\tan \left( {{\tan }^{-1}}x \right)$
But we know that ${{\tan }^{-1}}\left( \tan x \right)=\tan \left( {{\tan }^{-1}}x \right)=x$ hence we get,
$\Rightarrow \tan y=x$
Now we want to convert tan into sec hence we will try to bring it in the form ${{\tan }^{2}}y+1$
Now squaring on both sides we get,
$\Rightarrow {{\tan }^{2}}y={{x}^{2}}$
Now adding 1 on both sides we get,
$\Rightarrow {{\tan }^{2}}y+1={{x}^{2}}+1$
Now we know that $1+{{\tan }^{2}}x={{\sec }^{2}}x$ hence using this we get,
$\Rightarrow {{\sec }^{2}}y={{x}^{2}}+1$
Now taking square root on both sides we get,
$\Rightarrow \sec y=\sqrt{{{x}^{2}}+1}$
Now let us re-substitute the value of y. Hence substituting $y={{\tan }^{-1}}x$ we get,
$\Rightarrow \sec \left( {{\tan }^{-1}}\left( x \right) \right)=\sqrt{{{x}^{2}}+1}$

Note: We can also solve the equation without using the identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$ . Consider the equation ${{\tan }^{2}}y+1={{x}^{2}}+1$ converting tan in sin and cos we get, $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}+1=\sqrt{{{x}^{2}}+1}$ .
Now taking LCM and then using ${{\sin }^{2}}+{{\cos }^{2}}=1$ we will get the required equation. Also note that inverse functions are just represented by ${{f}^{-1}}$ and are not equal to $\dfrac{1}{f}$.