Question

How do you prove ${\sec ^2}x - {\tan ^2}x = 1$?

Answer 1

Hint: In order to proof the above statement ,take the left hand side of the equation and put ${\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}},{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}},$.now taking LCM and combining terms ,you will get \[1 - {\sin ^2}x\]in the numerator put it equal to ${\cos ^2}x$ according to the identity \[{\sin ^2}x + {\cos ^2}x = 1\] ,then simplifying further will give your final result which is equal to right-hand side of the equation.

Complete step by step answer:
To prove: ${\sec ^2}x - {\tan ^2}x = 1$
Proof: Taking Left-hand Side of the equation,
$\Rightarrow {\sec ^2}x - {\tan ^2}x$
As we know that $\tan x$ is equal to the ratio of $\sin x$ to $\cos x$ In simple words, $\tan x = \dfrac{{\sin x}}{{\cos x}},$ and if we square on both sides of this rule we get ${\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}},$ and $\sec x$ is the reciprocal of $\cos x$i.e. ${\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}$
Putting these values in the above equation, we get
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\]
As we can see the denominator of both of the terms is same , so we can directly add the numerator
\[ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}}\]
Using identity of trigonometry ,sum of square of sine and square of cosine is equal to one i.e. \[{\sin ^2}x + {\cos ^2}x = 1\].Rewriting it as \[{\cos ^2}x = 1 - {\sin ^2}x\].Putting this value in above equation we get
\[
\Rightarrow \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} \\
\Rightarrow 1 \\ \]
$\therefore LHS = 1$
Taking Right-hand Side part of the equation
$RHS = 1$
$\therefore LHS = RHS$
Hence, proved.

Additional Information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Even Function: A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.
3. Odd Function: A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $.Therefore,$\sin \theta $ and $\tan \theta $ and their reciprocals,$\cos ec\theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.
4. Periodic Function: A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.

Note:One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.Formula should be correctly used at every point.