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Question: Prove that \[\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\]...

Prove that cosxsinxcosx+sinx=cos2x1+sin2x\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}

Explanation

Solution

The given question deals with proving trigonometric equality using the basic and simple trigonometric formulae and identities such as cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x. So, we will first multiply the numerator and denominator of the left side of the result with cosx+sinx\cos x + \sin x. Then, we will evaluate the whole square of cosx+sinx\cos x + \sin x in the denominator and will use the double angle formula for cosine in the numerator to get the required result.

Complete step by step answer:
Given, we have to prove cosxsinxcosx+sinx=cos2x1+sin2x\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}
We know, the formulas that,
sin2x=2sinxcosx\sin 2x = 2\sin x\cos x
cos2x=cos2xsin2x\Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x
tan2x=2tanx1tan2x\Rightarrow \tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}},
Using these basic identities and some other methods we can prove the given question.

At first multiplying cosx+sinx\cos x + \sin xto both numerator and denominator of left hand side, we get
(cosxsinx)(cosx+sinx)(cosx+sinx)(cosx+sinx)\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\left( {\cos x + \sin x} \right)\left( {\cos x + \sin x} \right)}}
Now, using the formula (ab)(a+b)=a2b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}and (a+b)(a+b)=(a+b)2\left( {a + b} \right)\left( {a + b} \right) = {\left( {a + b} \right)^2}, we get
\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\left( {\cos x + \sin x} \right)\left( {\cos x + \sin x} \right)}}$$$$ = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}

Then using the formulas of double angle identities,
cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x, we get
cos2xsin2x(cosx+sinx)2=cos2x(cosx+sinx)2\Rightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}} = \dfrac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}
Now expanding the denominator in the formula of (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, we get
cos2xcos2x+2sinxcosx+sin2x\dfrac{{\cos 2x}}{{{{\cos }^2}x + 2\sin x\cos x + {{\sin }^2}x}}
We know that {\cos ^2}x$$$$ + {\sin ^2}x$$$$ = 1, so using this formula, we get,
cos2x1+2sinxcosx\dfrac{{\cos 2x}}{{1 + 2\sin x\cos x}}
Now from the formula of double angle identities, sin2x=2sinxcosx\sin 2x = 2\sin x\cos x, we get
cos2x1+sin2x\dfrac{{\cos 2x}}{{1 + \sin 2x}}

Thus cosxsinxcosx+sinx=cos2x1+sin2x\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}(proved).

Note: There are many forms of cosine double angle formulae such as cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x, cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1, cos2x=12sin2x\cos 2x = 1 - 2{\sin ^2}x, cos2x=(1tan2x1+tan2x)\cos 2x = \left( {\dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} \right). Hence, it is very crucial to understand which formula should be applied in the given problem. Since, we get, cos2xsin2x{\cos ^2}x - {\sin ^2}x in the numerator after multiplying and dividing the left side by cosx+sinx\cos x + \sin x, hence we apply cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x to simplify the expression. We should also remember the double angle formula for sine as sin2x=2sinxcosx\sin 2x = 2\sin x\cos x to solve the given problem.