Question

Find the range of $f(x)={{\csc }^{2}}x+25{{\sec }^{2}}x$.

Answer 1

Hint:In the given function apply basic trigonometric identities and convert it to tangent and cotangent functions. Now using the basic identity of ${{\left( a-b \right)}^{2}}$, simplify the expression that is given to us. We get $f(x)$ as a function which is greater than zero. Thus form the range of the function.

Complete step-by-step answer:
The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain. The range is the resulting y-values we get after substituting all the possible x-values. The range of a function is the spread of possible y-values (minimum y-value to maximum y-value)
Now we have been given the function,$f(x)={{\csc }^{2}}x+25{{\sec }^{2}}x$.
We know the basic trigonometric formulas like,

& {{\csc }^{2}}x=1+{{\cot }^{2}}x \\\ & {{\sec }^{2}}x=1+{{\tan }^{2}}x \\\ \end{aligned}$$ Let us substitute these values in the given function $$f(x)$$. Thus our expression becomes, $$\begin{aligned} & f(x)={{\csc }^{2}}x+25{{\sec }^{2}}x \\\ & f(x)=\left( 1+{{\cot }^{2}}x \right)+25\left( 1+{{\tan }^{2}}x \right) \\\ \end{aligned}$$ Now let us simplify it, $$\begin{aligned} & f(x)=\left( 1+{{\cot }^{2}}x \right)+25\left( 1+{{\tan }^{2}}x \right) \\\ & f(x)=1+{{\cot }^{2}}x+25+25{{\tan }^{2}}x \\\ & f(x)=26+{{\cot }^{2}}x+25{{\tan }^{2}}x \\\ \end{aligned}$$ Now let us add and subtract 10 from the above expression. $$\begin{aligned} & f(x)=26+{{\cot }^{2}}x+25{{\tan }^{2}}x-10+10 \\\ & f(x)=36+{{\cot }^{2}}x+25{{\tan }^{2}}x-10 \\\ \end{aligned}$$ Let us make the function in the form of $${{\left( a-b \right)}^{2}}$$. We know that, $$\begin{aligned} & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\\ & a=5\tan x,b=\cot x \\\ \end{aligned}$$ Hence we can write our expression as, $$\begin{aligned} & f(x)=36+{{\cot }^{2}}x+25{{\tan }^{2}}x-10 \\\ & f(x)=36+{{\left( 5\tan x \right)}^{2}}+{{\left( \cot x \right)}^{2}}-2\times \cot x\times 5\tan x \\\ & \because {{\left( 5\tan x \right)}^{2}}+{{\left( \cot x \right)}^{2}}-2\times \cot x\times 5\tan x={{\left( 5\tan x-\cot x \right)}^{2}} \\\ & \therefore f(x)=36+{{\left( 5\tan x-\cot x \right)}^{2}} \\\ \end{aligned}$$ Thus we got our function as, $$f(x)=36+{{\left( 5\tan x-\cot x \right)}^{2}}$$ Hence we can understand from this that, $${{\left( 5\tan x-\cot x \right)}^{2}}\ge 0$$ Now add 36 on both sides of the above expression to make it similar to our function $$f(x)$$. $${{\left( 5\tan x-\cot x \right)}^{2}}+36\ge 36$$ Thus we can say that, $$f(x)\ge 36$$. Thus from this we got the range of our function $$f(x)$$as, $$f(x)=\left[ 36,\infty \right)$$. Hence we got the range of the function $$f(x)={{\csc }^{2}}x+25{{\sec }^{2}}x$$ as $$\left[ 36,\infty \right)$$ . Note: You can even substitute $$f(x)$$as $$y$$, which is much easier to identify. By seeing the question, you should be able to identify which property you should be using in order to simplify it. It is important that you should know the range and domain of different functions such as $$\sin x,\cos x,\tan x$$etc, so that you might be able to find the range of functions related to it.