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# Let $f\left( x \right)={{\sec }^{-1}}x+{{\tan }^{-1}}x$, then $f\left( x \right)$ is real for:A. $x\in \left[ -1,1 \right]$ B. $x\in R$ C. $x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$ D. None of these

Last updated date: 16th Jul 2024
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Hint: For solving this question you should know about finding the values of inverse trigonometric values of the given functions. But if it is given to find the value of a trigonometric function, then we find the trigonometric value of that function and then we add them according to the given sequence.

According to our question t is asked of us to find the correct option for the value of $f\left( x \right)={{\sec }^{-1}}x+{{\tan }^{-1}}x$. As we know that if any question is asking for the values of trigonometric functions, then we directly put the value of that function there and get our required answer for that. But if it is given to find the angles of them, then we find the original function (trigonometric function) for that value and then we get the angle of that by comparing it with each other. And if it is asked directly without any angle, then we always use the general formulas of trigonometry for solving this. And we will solve that according to the given sequence of calculations.
$f\left( x \right)={{\sec }^{-1}}x+{{\tan }^{-1}}x$
And the value of $\sec x$ and $\operatorname{cosec}x$ does not lie between +1 and -1.
If $\phi \left( x \right)={{\sec }^{-1}}x$, then we know that, $x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$.
Also $g\left( x \right)={{\tan }^{-1}}x$, then $x\in R$.
Hence for holding $f\left( x \right)={{\sec }^{-1}}x+{{\tan }^{-1}}x$, we have $x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$.