Question

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

Answer 1

Hint: In this particular type of question use the concept that consider the L.H.S part or the R.H.S part, if we consider the L.H.S part then use the property that (${\sec ^2}x = 1 + {\tan ^2}x$) and use that (${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$) so first simplify according to trigonometric property then simplify the above basic identity so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given trigonometric equation
${\sec ^4}x - {\sec ^2}x = {\tan ^4}x + {\tan ^2}x$
Consider the L.H.S part of the equation we have,
$ \Rightarrow {\sec ^4}x - {\sec ^2}x$
This equation is also written as
$ \Rightarrow {\left( {{{\sec }^2}x} \right)^2} - {\sec ^2}x$
Now as we know that (${\sec ^2}x = 1 + {\tan ^2}x$) so use this property in the above equation we have,
$ \Rightarrow {\left( {1 + {{\tan }^2}x} \right)^2} - \left( {1 + {{\tan }^2}x} \right)$
Now open the square according to property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ so we have,
$ \Rightarrow \left( {1 + {{\tan }^4}x + 2{{\tan }^2}x} \right) - \left( {1 + {{\tan }^2}x} \right)$
Now simplify it we have,
$ \Rightarrow 1 + {\tan ^4}x + 2{\tan ^2}x - 1 - {\tan ^2}x$
$ \Rightarrow {\tan ^4}x + 2{\tan ^2}x - {\tan ^2}x$
\[ \Rightarrow {\tan ^4}x + {\tan ^2}x\]
= R.H.S
Hence proved.

Note: In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions . Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar trigonometric functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then it is the required answer.