Question

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

Answer 1

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.