Question

Prove the given trigonometric equation,
 ${{\sec }^{4}}x-{{\sec }^{2}}x$ = ${{\tan }^{4}}x+{{\tan }^{2}}x$.

Answer 1

Hint: we will use Trigonometric identities like ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. Few other identities which can be remembered are.
 \[\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & {{\csc }^{2}}x-{{\cot }^{2}}x=1 \\
 & \sec x=\frac{1}{\cos x} \\
 & \csc x=\frac{1}{\sin x} \\
\end{align}\]

Complete step-by-step answer:

We will be taking the Left-hand side of the equation and proving it equal to the right-hand side.
So, by taking LHS we have
${{\sec }^{4}}x-{{\sec }^{2}}x$
Now we will take ${{\sec }^{2}}x$ common in the above expression
=${{\sec }^{2}}x({{\sec }^{2}}x-1)$
We know that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. We can rewrite it as
${{\sec }^{2}}x-1={{\tan }^{2}}x$ ….…... (1)
Or
${{\sec }^{2}}x=1+{{\tan }^{2}}x$ ….….…. (2)

We will substitute (1) and (2) in ${{\sec }^{2}}x({{\sec }^{2}}x-1)$ and as a resultant we get
=$(1+{{\tan }^{2}}x){{\tan }^{2}}x$
= ${{\tan }^{2}}x+{{\tan }^{4}}x$
Since the left side of the equation on simplifying comes equal to the right-hand side. Hence proved.

Note: Alternative way
We can solve it by using these identities
\[\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1.......(1) \\
 & {{\csc }^{2}}x-{{\cot }^{2}}x=1........(2) \\
 & \sec x=\dfrac{1}{\cos x}........(3) \\
 & \csc x=\dfrac{1}{\sin x}.........(4) \\
\end{align}\]

So, we can write ${{\sec }^{4}}x-{{\sec }^{2}}x$as
${{\sec }^{2}}x({{\sec }^{2}}x-1)$ by taking ${{\sec }^{2}}x$as common.
Now, we will write $\sec x$ in its reverse form by using the identity (3) doing this we will convert the whole equation in the form of sine and cosine.
$\begin{align}
  & =\dfrac{1}{{{\cos }^{2}}x}(\dfrac{1-{{\cos }^{2}}x}{{{\cos }^{2}}x}) \\
 & \\
\end{align}$ ………. (5)
Now since we have converted the whole equation in the form of sine and cosine we will use the relation between sine and cosine to further simplify the equation by using (1).
Now using (1) we can write
$1-{{\cos }^{2}}x={{\sin }^{2}}x$
Now we will substitute this in (5) to further simplify the expression.
=$\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x}(\dfrac{\sin x}{\cos x})$
Now since we know that $\dfrac{\sin x}{\cos x}=\tan x$ we use it in the above equation.
=$(\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}+1)(\tan x)$
Now again using $\dfrac{\sin x}{\cos x}=\tan x$ we can write
=$({{\tan }^{2}}x+1)(\tan x)$
=${{\tan }^{4}}x+{{\tan }^{2}}x$
Hence proved