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

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

Explanation

Solution

In order to proof the above statement ,take the left hand side of the equation and expanding the whole square by using the formula (A−B)2=A2+B2−2.A.B{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B by considering A =1 and B=tan⁡xA\, = 1\,and\,B = \tan x .Simplifying further using identity of trigonometry sec⁡2x=tan⁡2x+1{\sec ^2}x = {\tan ^2}x + 1,you will give your final result which is equal to right-hand side of the equation.

Complete step by step answer:
To prove: (1−tan⁡x)2=sec⁡2x−2tan⁡x{\left( {1 - \tan x} \right)^2} = {\sec ^2}x - 2\tan x
Proof: Taking Left-hand Side of the equation,
⇒(1−tan⁡x)2\Rightarrow {\left( {1 - \tan x} \right)^2}
Now expanding the above equation by Using the formula of square of difference of two numbers (A−B)2=A2+B2−2.A.B{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B by considering A =1 and B=tan⁡xA\, = 1\,and\,B = \tan x.Our equation now becomes
⇒12+(tan⁡x)2−2(1)(tan⁡x)\Rightarrow {1^2} + {\left( {\tan x} \right)^2} - 2\left( 1 \right)\left( {\tan x} \right)
Simplifying further, we get
⇒1+tan⁡2x−2tan⁡x\Rightarrow 1 + {\tan ^2}x - 2\tan x
Using identity of trigonometry sec⁡2x=tan⁡2x+1{\sec ^2}x = {\tan ^2}x + 1.Putting this in the above , we get
⇒sec⁡2x−2tan⁡x\Rightarrow {\sec ^2}x - 2\tan x
∴LHS=sec⁡2x−2tan⁡x\therefore LHS = {\sec ^2}x - 2\tan x
Taking Right-hand Side part of the equation
RHS=sec⁡2x−2tan⁡xRHS = {\sec ^2}x - 2\tan x
∴LHS=RHS\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)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
3. Odd Function: A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta .Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals,cos⁡ecθ\cos ec\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.
4. Periodic Function: A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)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.