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Question: How do you prove trigonometric identity \(1 + {\tan ^2}x = {\sec ^2}x\)?...

How do you prove trigonometric identity 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x?

Explanation

Solution

This question is from the topic of trigonometric formulas. In this question we have to prove trigonometric identity 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x. To prove this we will use the definition of tangent and secant trigonometric functions and trigonometric formula sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1. We prove this by starting with L.H.S and getting R.H.S by using properties or formulas.

Complete step by step solution:
Let us try to solve the question in which we are asked to prove 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x. To solve these types of questions we need to know the definition and trigonometric formulas without them it is impossible to solve these questions. To prove this we will first use some of this property.
Here are those properties of trigonometric functions:
3) tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}}
4) secx=1cosx\sec x = \dfrac{1}{{\cos x}}
And use the trigonometric formula sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1to prove the given trigonometric formula 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x.
To prove: 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x
Proof: To prove the above equation we will start with the L.H.S of equation.
So we have,
L.H.S=1+tan2xL.H.S = 1 + {\tan ^2}x eq(1)eq(1)
As we know that tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}}, so we have tan2x=sin2xcos2x{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} putting this back into eq(1)eq(1), we get
L.H.S=1+sin2xcos2xL.H.S = 1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}
Now performing fraction addition we get,
L.H.S=1+sin2xcos2x L.H.S=cos2x+sin2xcos2x  L.H.S = 1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} \\\ L.H.S\,\, = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} \\\
Now using the trigonometric formula sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1. We get,
L.H.S=1cos2xL.H.S = \dfrac{1}{{{{\cos }^2}x}}
Now using property 2 from above which is secx=1cosx\sec x = \dfrac{1}{{\cos x}}. We get,
L.H.S=sec2xL.H.S = {\sec ^2}x
1+tan2x=L.H.S=R.H.S=sec2x1 + {\tan ^2}x = L.H.S = R.H.S = {\sec ^2}x
Hence we have proved that1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x.

Note: Generally questions in which we are asked to prove a trigonometric formula from L.H.S of equation but this can be proved also starting with R.H.S. In proving the above formula we just have to apply reverse steps of the above proof. You can try to prove it by starting with sec2x{\sec ^2}x. For solving this type of question you are required to know the trigonometric formulas.