Question

Find the range of the trigonometric function $f(x)=\sqrt{4-\sqrt{1+{{\tan }^{2}}x}}$

Answer 1

Hint: In this question, you can use the trigonometric (Pythagorean) identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$ and you know that, the trigonometric secant function $\left| \sec x \right|\ge 1$.

Complete step-by-step answer:
The given trigonometric function is

$f(x)=\sqrt{4-\sqrt{1+{{\tan }^{2}}x}}$

We know that, trigonometric (Pythagorean) identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$

$f(x)=\sqrt{4-\sqrt{{{\sec }^{2}}x}}$

Now $\sqrt{{{\sec }^{2}}x}=\left| \sec x \right|$

$f(x)=\sqrt{4-\left| \sec x \right|}$

We know that, the trigonometric secant function$\left| \sec x \right|\ge 1$.

$\left| \sec x \right|\ge 1$

Multiplying both sides by -1, we get

$-\left| \sec x \right|\le -1$

Adding 4 on the both sides, we get

$4-\left| \sec x \right|\le 4-1$

\[4-\left| \sec x \right|\le 3\]

Taking square root on the both sides, we get

\[\sqrt{4-\left| \sec x \right|}\le \sqrt{3}\]

For the given function,, the minimum value is zero.

\[0\le \sqrt{4-\left| \sec x \right|}\le \sqrt{3}\]

Now put $f(x)=\sqrt{4-\left| \sec x \right|}$, we get

\[0\le f(x)\le \sqrt{3}\]

Hence the range of the given function is $\left[ 0,\sqrt{3} \right]$.


Note: You might get confused about the difference between the domain of a function and the range of a function. Domain is the independent variable and range is the dependent variable. On the other hand, range is defined as a set of all probable output values. Domain is what is put into a function, whereas range is what is the result of the function with the domain value.