Question

What does $\sec x$ equal in terms of $\sin $,$\cos $ and/or $\tan $?

Answer 1

Hint: In this question we have been given the trigonometric identity $\sec x$ and we have to express it in the form of the trigonometric identities of $\sin ,\cos $ and $\tan $. We will use the properties of trigonometric identities and write the expression $\sec x$ in the terms of the given functions. We will use the Pythagorean identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and substitute it to get the required solution.

Complete step-by-step solution:
We have the term given to us as:
$\Rightarrow \sec x$
We know that $\sec x$ is the inverse of $\cos x$ which means that $\cos x=\dfrac{1}{\sec x}$.
On rearranging the terms, we get:
$\Rightarrow \sec x=\dfrac{1}{\cos x}\to \left( 1 \right)$
Which is the required value of $\sec x$ in terms of $\cos x$.
Now we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, on rearranging, we get:
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
On taking the square root on both the sides, we get:
$\Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x}\to \left( 2 \right)$
On substituting the value of equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get:
$\Rightarrow \sec x=\dfrac{1}{\sqrt{1-{{\sin }^{2}}x}}$
Which is the required value of $\sec x$ in terms of $\sin x$.
Now we know the Pythagorean identity as:
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1$
On dividing both the sides of the expression by ${{\cos }^{2}}x$, we get:
$\Rightarrow \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$
On splitting the denominator in the left-hand side of the expression, we get:
$\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}+\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$
Now we know that $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{1}{\cos x}=\sec x$ therefore, on substituting, we get:
$\Rightarrow {{\tan }^{2}}x+1={{\sec }^{2}}x$
On rearranging the expression, we get:
$\Rightarrow {{\sec }^{2}}x=1+{{\tan }^{2}}x$
On taking the square root on both the sides, we get:
$\Rightarrow \sec x=\sqrt{1+{{\tan }^{2}}x}$
Which is the required value of $\sec x$ in terms of $\tan x$.
Therefore, we get the solution as:
$\Rightarrow \sec x=\dfrac{1}{\cos x}$
$\Rightarrow \sec x=\dfrac{1}{\sqrt{1-{{\sin }^{2}}x}}$
$\Rightarrow \sec x=\sqrt{1+{{\tan }^{2}}x}$

Note: It is to be remembered there also exists other trigonometric quantities such as $\csc x$ and $\cot x$. $\csc x$ is the inverse function of $\sin x$ and $\cot x$ is the inverse function of $\tan x$. The correlation between the various trigonometric functions should be remembered and the Pythagoras rule should also be remembered for sums which require its substitutions.