Question

Prove ${\tan ^4}x + 2{\tan ^2}x + 1 = {\sec ^4}x$.

Answer 1

Hint: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 one of the basic trigonometric identities given below, we can simplify the above expression.
$1 + {\tan ^2}x = {\sec ^2}x$
In order to verify the given expression we have to use the above identity and express our given expression in that form and thereby verify it.

Complete step by step answer:
Given, ${\tan ^4}x + 2{\tan ^2}x + 1 = {\sec ^4}x.............................\left( i \right)$.
Now in order to prove (i) 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:
\[{\tan ^4}x + 2{\tan ^2}x + 1 \\
\Rightarrow{\tan ^4}x + {\tan ^2}x + {\tan ^2}x + 1.......................\left( {ii} \right) \\ \]
Now we have the identity $1 + {\tan ^2}x = {\sec ^2}x$
The above identity can be substituted in (ii) such that we get:
\[{\tan ^4}x + {\tan ^2}x + {\tan ^2}x + 1 = {\tan ^4}x + {\tan ^2}x + {\sec ^2}x.............\left( {iii} \right)\]
Now on observing the \[{\tan ^4}x + {\tan ^2}x\] terms in (iii) we can say that the term \[{\tan ^2}x\] is common. Such that:
\[{\tan ^4}x + {\tan ^2}x + {\sec ^2}x = {\tan ^2}x\left( {{{\tan }^2}x + 1} \right) + {\sec ^2}x.....................\left( {iv} \right)\]
Now again apply the identity $1 + {\tan ^2}x = {\sec ^2}x$ in the above equation.Such that:
\[{\tan ^2}x\left( {{{\tan }^2}x + 1} \right) + {\sec ^2}x = {\tan ^2}x{\sec ^2}x + {\sec ^2}x....................\left( v \right)\]
Now on observing the \[{\tan ^2}x{\sec ^2}x + {\sec ^2}x\] terms in (iii) we can say that the term \[{\sec ^2}x\] is common. Such that:
\[{\tan ^2}x{\sec ^2}x + {\sec ^2}x = {\sec ^2}x\left( {{{\tan }^2}x + 1} \right)..........................\left( {vi} \right)\]
Now again apply the identity $1 + {\tan ^2}x = {\sec ^2}x$ in the above equation.
Such that:
\[{\sec ^2}x\left( {{{\tan }^2}x + 1} \right) = {\sec ^2}x \times {\sec ^2}x \\
\therefore{\sec ^2}x\left( {{{\tan }^2}x + 1} \right) = {\sec ^4}x.........................\left( {vii} \right) \\ \]
Therefore from (viii) we can say that LHS =RHS.

Hence Proved ${\tan ^4}x + 2{\tan ^2}x + 1 = {\sec ^4}x$

Note:Some other equations needed for solving these types of problem are:
\[\sec x = \dfrac{1}{{\cos x}} \\
\tan x = \dfrac{{\sin x}}{{\cos x}} \\
{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\
{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin }^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1} \]
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.