Question

How do you simplify $$\sec (\arctan (4x))$$?

Answer 1

Hint: We will consider the inner function as an angle. Then we will eliminate the function $$\arctan$$ from the resulting expression. Next, we will draw a right triangle based on the expression. Finally, we will find $$\sec$$ of the angle we have considered.

Formula used:
 $$\tan (\arctan (x))=x$$

Complete step by step solution:
We are required to simplify the expression $$\sec (\arctan (4x))$$.
Let us begin by denoting the innermost function, which is $$\arctan (4x)$$ as an angle $$\theta$$. We get
$$\theta =\arctan (4x)$$ ………$$(1)$$
Let us try to eliminate the function $$\arctan$$ from equation $$(1)$$ by taking $$\tan$$ on both sides of equation $$(1)$$. This gives us
$$\tan \theta =\tan (\arctan (4x))$$ ………..$$(2)$$
We will use the property $$\tan (\arctan (x))=x$$ on the RHS of equation $$(2)$$ to get the following expression:
$$\tan \theta =4x$$ ………$$(3)$$
Let us consider a right triangle ABC with right-angle $$\mathrm{\angle }B$$ as in the figure. Let us take $$\theta$$ as the angle between the sides $$AC$$ and $$BC$$.

Now, from equation $$(3)$$, we can write
$$\tan \theta ={\displaystyle \frac{4x}{1}}$$
We know that in a right-triangle ABC,
$$\tan \theta ={\displaystyle \frac{\mathrm{o}\mathrm{p}\mathrm{p}.\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}}{\mathrm{a}\mathrm{d}\mathrm{j}.\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}}}$$. Compared with the above equation, we can take the opposite side of the right-triangle as $$4x$$ units and the adjacent side as $$1$$ unit.
Now, we have to find the hypotenuse of the right triangle with sides $$4x$$ units and $$1$$ units.
By Pythagoras theorem, we have $$A{C}^2=A{B}^2+B{C}^2$$. Substituting $$AC=y,AB=4x$$ and $$BC=1$$, we get
$$y^2=(4x{)}^2+1^2$$
Taking square root on both sides of the above expression, we get
$$y=\sqrt{16x^2+1}$$ ……..$$(4)$$
We are supposed to find the value of $$\sec (\arctan (4x))$$, which from equation $$(1)$$ is the same as $$\sec \theta$$.
We know that $$\sec \theta ={\displaystyle \frac{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{u}\mathrm{s}\mathrm{e}}{\mathrm{a}\mathrm{d}\mathrm{j}.\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}}}$$. So, from triangle ABC and equation $$(4)$$, we get
$$\sec \theta ={\displaystyle \frac{\sqrt{16x^2+1}}{1}}=\sqrt{16x^2+1}$$

Note:
The functions $$\tan$$ and $$\arctan$$ are inverse functions. We have used the property $$\tan (\arctan (x))=x$$ to eliminate the function $$\arctan$$, since we cannot use that function in a right-angled triangle. Similarly, we have properties $$\sin (\arcsin (x))=x$$ and $$\cos (\arccos (x))=x$$.