Question

Verify ${\sec ^4}x - {\tan ^4}x = {\sec ^2}x + {\tan ^2}x.$

Answer 1

Now in order to verify the above statement we should work with one side at a time and manipulate it to the other side. Using some basic trigonometric identities like we can simplify the above expression.
\sec x = \dfrac{1}{{\cos x}} \\\ \Rightarrow\tan x = \dfrac{{\sin x}}{{\cos x}} \\\
Such that in order to verify the given expression we have to use the above identities and express our given expression in that form and thereby verify it.

Complete step by step answer:
Given, ${\sec ^4}x - {\tan ^4}x = {\sec ^2}x + {\tan ^2}x.............................\left( i \right)$
Now in order to verify it we have to simplify either the LHS or the RHS of the equation towards RHS or the LHS of the equation respectively.Here let’s take the LHS of the equation which is:
{\sec ^4}x - {\tan ^4}x \\\ \Rightarrow{\sec ^4}x - {\tan ^4}x = \left( {{{\sec }^2}x + {{\tan }^2}x} \right)\left( {{{\sec }^2}x - {{\tan }^2}x} \right)............\left( {ii} \right) \\\
Now since in the RHS we have $\left( {{{\sec }^2}x + {{\tan }^2}x} \right)$let’s simplify the term $\left( {{{\sec }^2}x - {{\tan }^2}x} \right)$and substitute in (iii) since our aim is to verify ${\sec ^4}x - {\tan ^4}x = {\sec ^2}x + {\tan ^2}x.$
Let’s simplify the term $\left( {{{\sec }^2}x - {{\tan }^2}x} \right).......................\left( {iii} \right)$
Now we know that:
\sec x = \dfrac{1}{{\cos x}} \\\ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}} \\\
Substituting these values in (iii) we get:

{\sec ^2}x - {\tan ^2}x = \dfrac{1}{{{{\cos }^2}x}} - {\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2} \\\ \Rightarrow {\sec ^2}x - {\tan ^2}x = \dfrac{1}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}..........................\left( {iv} \right) \\\ $$ Now since in (iv) the denominator is same we can perform direct subtraction such that: $\dfrac{1}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}}...................\left( v \right)$ We also know the equation: $ {\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow 1 - {\sin ^2}x = {\cos ^2}x..................\left( {vi} \right) \\\ $ Substituting (vi) in (v): $\dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = 1.......\left( {vii} \right)$ So we can say that: $$\left( {{{\sec }^2}x - {{\tan }^2}x} \right) = 1......................\left( {viii} \right)$$ Now substitute (viii) in (ii) we get: $ \Rightarrow {\sec ^4}x - {\tan ^4}x = \left( {{{\sec }^2}x + {{\tan }^2}x} \right)\left( {{{\sec }^2}x - {{\tan }^2}x} \right) \\\ \therefore{\sec ^4}x - {\tan ^4}x = \left( {{{\sec }^2}x + {{\tan }^2}x} \right).............................\left( {ix} \right) \\\ $ Therefore from (ix) we can say that LHS =RHS. **Hence verified, ${\sec ^4}x - {\tan ^4}x = {\sec ^2}x + {\tan ^2}x.$** **Note:** Some other equations needed for solving these types of problem are:

1 + {\tan ^2}x = {\sec ^2}x \\
\begin{array}{*{20}{l}}
\Rightarrow{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\
\Rightarrow{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin }^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1}
\end{array} \\ $$
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.