Question

Is $ {\tan ^2}x = {\sec ^2}x - 1 $ an identity?

Answer 1

Hint: There are three main identities in trigonometry.
 $
   \Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\
   \Rightarrow 1 + {\tan ^2}x = {\sec ^2}x \\
   \Rightarrow 1 + {\cot ^2}x = \cos e{c^2}x \;
  $
Using relations and mathematical calculations, we can derive these identities in many different forms. To check whether the given form is derived from one of these, we will compare the given form with the main form and see if it can be derived back to the main form or not.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we are supposed to check if it is an identity or not.
Given equation: $ {\tan ^2}x = {\sec ^2}x - 1 $
Now, we know the three famous trigonometric identities are:
 $
   \Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\
   \Rightarrow 1 + {\tan ^2}x = {\sec ^2}x \\
   \Rightarrow 1 + {\cot ^2}x = \cos e{c^2}x \;
  $
Now, if take the identity $ 1 + {\tan ^2}x = {\sec ^2}x $ and subtract 1 on both LHS and RHS, we will get
 $
   \Rightarrow 1 + {\tan ^2}x - 1 = {\sec ^2}x - 1 \\
   \Rightarrow {\tan ^2}x = {\sec ^2}x - 1 \;
  $
So, we can say that the given equation is a trigonometric identity.
Now, let us see how it was derived.
So, the given equation is $ {\tan ^2}x = {\sec ^2}x - 1 $ .
Let us take the RHS of the given equation.
 $ \Rightarrow RHS = {\sec ^2}x - 1 $ - - - - - - - - - (1)
Now, we know that sec is the inverse of cos. So, we can write sec as 1 divided by cos. Therefore, equation (1) becomes,
 $ \Rightarrow RHS = \dfrac{1}{{{{\cos }^2}x}} - 1 $
Now, take LCM
 $ \Rightarrow RHS = \dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}} $ - - - - - - - - - - - (2)
Now, we have another identity $ {\sin ^2}x + {\cos ^2}x = 1 $ . So, if we subtract $ {\cos ^2}x $ on both LHS and RHS, we will get
 $
   \Rightarrow {\sin ^2}x + {\cos ^2}x - {\cos ^2}x = 1 - {\cos ^2}x \\
   \Rightarrow 1 - {\cos ^2}x = {\sin ^2}x \;
  $
Substituting this value in equation (2), we get
 $ \Rightarrow RHS = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} $
Now, we know that $ \dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ . Therefore,
 $ \Rightarrow RHS = {\tan ^2}x $
 $ \Rightarrow RHS = LHS $
Hence, the given identity is correct and we have seen its derivation.

Note: We can also prove the other identity $ 1 + {\cot ^2}x = \cos e{c^2}x $ .
Here, take the LHS of the equation, we get
 $ LHS = 1 + {\cot ^2}x $
Now, $ \cot x = \dfrac{{\cos x}}{{\sin x}} $
 $ LHS = 1 + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} $
           $ = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} $
Now, we know that $ {\sin ^2}x + {\cos ^2}x = 1 $
 $ LHS = \dfrac{1}{{{{\sin }^2}x}} = \cos e{c^2}x $
Hence, $ LHS = RHS $ .