Solveeit Logo

Question

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

How do you prove (1tanx)2=sec2x2tanx{\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 (AB)2=A2+B22.A.B{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B by considering A=1andB=tanxA\, = 1\,and\,B = \tan x .Simplifying further using identity of trigonometry sec2x=tan2x+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: (1tanx)2=sec2x2tanx{\left( {1 - \tan x} \right)^2} = {\sec ^2}x - 2\tan x
Proof: Taking Left-hand Side of the equation,
(1tanx)2\Rightarrow {\left( {1 - \tan x} \right)^2}
Now expanding the above equation by Using the formula of square of difference of two numbers (AB)2=A2+B22.A.B{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2.A.B by considering A=1andB=tanxA\, = 1\,and\,B = \tan x.Our equation now becomes
12+(tanx)22(1)(tanx)\Rightarrow {1^2} + {\left( {\tan x} \right)^2} - 2\left( 1 \right)\left( {\tan x} \right)
Simplifying further, we get
1+tan2x2tanx\Rightarrow 1 + {\tan ^2}x - 2\tan x
Using identity of trigonometry sec2x=tan2x+1{\sec ^2}x = {\tan ^2}x + 1.Putting this in the above , we get
sec2x2tanx\Rightarrow {\sec ^2}x - 2\tan x
LHS=sec2x2tanx\therefore LHS = {\sec ^2}x - 2\tan x
Taking Right-hand Side part of the equation
RHS=sec2x2tanxRHS = {\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)=sinθ.cos(θ)=cosθandtan(θ)=tanθ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta .Therefore, sinθ\sin \theta and tanθ\tan \theta and their reciprocals,cosecθ\cos ec\theta and cotθ\cot \theta are odd functions whereas cosθ\cos \theta and its reciprocal secθ\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.