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

Last updated date: 15th Jul 2024
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Hint: In order to proof the above statement ,take the left hand side of the equation and expanding the whole square by using the formula ${\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B$ by considering $A\, = 1\,and\,B = \tan x$ .Simplifying further using identity of trigonometry ${\sec ^2}x = {\tan ^2}x + 1$,you will give your final result which is equal to right-hand side of the equation.

To prove: ${\left( {1 - \tan x} \right)^2} = {\sec ^2}x - 2\tan x$
Proof: Taking Left-hand Side of the equation,
$\Rightarrow {\left( {1 - \tan x} \right)^2}$
Now expanding the above equation by Using the formula of square of difference of two numbers ${\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B$ by considering $A\, = 1\,and\,B = \tan x$.Our equation now becomes
$\Rightarrow {1^2} + {\left( {\tan x} \right)^2} - 2\left( 1 \right)\left( {\tan x} \right)$
Simplifying further, we get
$\Rightarrow 1 + {\tan ^2}x - 2\tan x$
Using identity of trigonometry ${\sec ^2}x = {\tan ^2}x + 1$.Putting this in the above , we get
$\Rightarrow {\sec ^2}x - 2\tan x$
$\therefore LHS = {\sec ^2}x - 2\tan x$
Taking Right-hand Side part of the equation
$RHS = {\sec ^2}x - 2\tan x$
$\therefore LHS = RHS$
Hence, proved.

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.