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Question: How do you prove \(\dfrac{{1 - \sin 2x}}{{\cos 2x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\)?...

How do you prove 1−sin⁡2xcos⁡2x=cos⁡2x1+sin⁡2x\dfrac{{1 - \sin 2x}}{{\cos 2x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}?

Explanation

Solution

We will use the double angle trigonometric formulas to simplify any one side of the equation or both sides so that the result for both should come same and therefore, we can prove that L.H.S. of the equation is equal to the R.H.S.

Complete step by step solution:
We have to prove 1−sin⁡2xcos⁡2x=cos⁡2x1+sin⁡2x\dfrac{{1 - \sin 2x}}{{\cos 2x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}} where
L.H.S. = 1−sin⁡2xcos⁡2x\dfrac{{1 - \sin 2x}}{{\cos 2x}} and
R.H.S. = cos⁡2x1+sin⁡2x\dfrac{{\cos 2x}}{{1 + \sin 2x}}
First, we will solve the L.H.S. of the equation,
L.H.S. = 1−sin⁡2xcos⁡2x\dfrac{{1 - \sin 2x}}{{\cos 2x}}
Expanding the identities sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x,
L.H.S. = 1−2sin⁡xcos⁡xcos⁡2x−sin⁡2x\dfrac{{1 - 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}
In the numerator, we write 1=cos⁡2x+sin⁡2x1 = {\cos ^2}x + {\sin ^2}x,
L.H.S. = cos⁡2x+sin⁡2x−2sin⁡xcos⁡xcos⁡2x−sin⁡2x\dfrac{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}
Using the formula, a2−2ab+b2=(a−b)2{a^2} - 2ab + {b^2} = {(a - b)^2}, we can write the numerator as,
L.H.S. = (cos⁡x−sin⁡x)2cos⁡2x−sin⁡2x\dfrac{{{{(\cos x - \sin x)}^2}}}{{{{\cos }^2}x - {{\sin }^2}x}}
Using the formula, a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b), we can write the denominator as,
L.H.S. = (cos⁡x−sin⁡x)2(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\dfrac{{{{(\cos x - \sin x)}^2}}}{{(\cos x - \sin x)(\cos x + \sin x)}}
Cancelling cos⁡x−sin⁡x\cos x - \sin x from both numerator and denominator,
L.H.S. = (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\dfrac{{(\cos x - \sin x)}}{{(\cos x + \sin x)}}
Multiplying and dividing equation by cos⁡x+sin⁡x\cos x + \sin x, we get,
L.H.S. = cos⁡x−sin⁡xcos⁡x+sin⁡x×cos⁡x+sin⁡xcos⁡x+sin⁡x\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \times \dfrac{{\cos x + \sin x}}{{\cos x + \sin x}}
Multiplying the numerator and denominator part,
L.H.S. = (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(cos⁡x+sin⁡x)(cos⁡x+sin⁡x)\dfrac{{(\cos x - \sin x)(\cos x + \sin x)}}{{(\cos x + \sin x)(\cos x + \sin x)}}
Using the formula, (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}in the numerator, we get,
L.H.S. = cos⁡2x−sin⁡2x(cos⁡x+sin⁡x)(cos⁡x+sin⁡x)\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{(\cos x + \sin x)(\cos x + \sin x)}}
Using the formula, a.a=a2a.a = {a^2} in the denominator, we get,
L.H.S. = cos⁡2x−sin⁡2x(cos⁡x+sin⁡x)2\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{(\cos x + \sin x)}^2}}}
Expanding the denominator using the formula (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2} ,
L.H.S. = cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x+2sin⁡xcos⁡x\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x + {{\sin }^2}x + 2\sin x\cos x}}
We know that, the trigonometric identity cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1, so substituting that,
L.H.S. = cos⁡2x−sin⁡2x1+2sin⁡xcos⁡x\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{1 + 2\sin x\cos x}}
The term cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x is a double angle formula of cos⁡2x\cos 2x, so substituting cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x, we get,
L.H.S. = cos⁡2x1+2sin⁡xcos⁡x\dfrac{{\cos 2x}}{{1 + 2\sin x\cos x}}
We know that sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x which is double angle formula, therefore, substituting this value,
L.H.S. = cos⁡2x1+sin⁡2x\dfrac{{\cos 2x}}{{1 + \sin 2x}} = R.H.S.
Therefore, we have proved L.H.S. = R.H.S.
Hence proved 1−sin⁡2xcos⁡2x=cos⁡2x1+sin⁡2x\dfrac{{1 - \sin 2x}}{{\cos 2x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}.

Note:
We can also start by simplifying the right-hand side with the same method as we did above. While simplifying trigonometric problems, one should have a proper knowledge of all the trigonometric formulas and identities and basic arithmetic formulas such as (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}, (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2} , etc.