Question
Question: Prove that \[\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\]...
Prove that cosx+sinxcosx−sinx=1+sin2xcos2x
Solution
The given question deals with proving trigonometric equality using the basic and simple trigonometric formulae and identities such as cos2x=cos2x−sin2x. So, we will first multiply the numerator and denominator of the left side of the result with cosx+sinx. Then, we will evaluate the whole square of cosx+sinx 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 cosx+sinxcosx−sinx=1+sin2xcos2x
We know, the formulas that,
sin2x=2sinxcosx
⇒cos2x=cos2x−sin2x
⇒tan2x=1−tan2x2tanx,
Using these basic identities and some other methods we can prove the given question.
At first multiplying cosx+sinxto both numerator and denominator of left hand side, we get
(cosx+sinx)(cosx+sinx)(cosx−sinx)(cosx+sinx)
Now, using the formula (a−b)(a+b)=a2−b2and (a+b)(a+b)=(a+b)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)}}$$$$ = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}
Then using the formulas of double angle identities,
cos2x=cos2x−sin2x, we get
⇒(cosx+sinx)2cos2x−sin2x=(cosx+sinx)2cos2x
Now expanding the denominator in the formula of (a+b)2=a2+2ab+b2, we get
cos2x+2sinxcosx+sin2xcos2x
We know that {\cos ^2}x$$$$ + {\sin ^2}x$$$$ = 1, so using this formula, we get,
1+2sinxcosxcos2x
Now from the formula of double angle identities, sin2x=2sinxcosx, we get
1+sin2xcos2x
Thus cosx+sinxcosx−sinx=1+sin2xcos2x(proved).
Note: There are many forms of cosine double angle formulae such as cos2x=cos2x−sin2x, cos2x=2cos2x−1, cos2x=1−2sin2x, cos2x=(1+tan2x1−tan2x). Hence, it is very crucial to understand which formula should be applied in the given problem. Since, we get, cos2x−sin2x in the numerator after multiplying and dividing the left side by cosx+sinx, hence we apply cos2x=cos2x−sin2x to simplify the expression. We should also remember the double angle formula for sine as sin2x=2sinxcosx to solve the given problem.