Question

Simplify $\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}$ .

Answer 1

Here we will first rationalise the given expression by multiplying and dividing with $\left( {\cos x - \sin x} \right)$ .
We will also use the algebraic formula of the difference between two terms: ${(a - b)^2} = {a^2} + {b^2} + 2ab$ .
Now we will use the fact that $\cos 2x = {\cos ^2}x - {\sin ^2}x$ , thus we get the required answer.

Complete step by step solution:
Here we have $\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}$ .
Let us first rationalise the expression i.e.
$\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \times \dfrac{{\cos x - \sin x}}{{\cos x - \sin x}}$ .
On multiplying we have
$\dfrac{{{{(\cos x - \sin x)}^2}}}{{{{\cos }^2}x - {{\sin }^2}x}}$
We will apply the square of the difference formula as mentioned in the hint,
${(a - b)^2} = {a^2} + {b^2} + 2ab$
By comparing with formula we have
$a = \cos x,b = \sin x$
So by using the difference formula in numerator, we can write
$\dfrac{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
We know the double angle formula of sine function which states:
$\sin 2x = 2\sin x\cos x$
Also we know the basic trigonometric identity
${\sin ^2}x + {\cos ^2}x = 1$
By substituting these values in the numerator we have:
$\dfrac{{1 - \sin 2x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
From the hint we can substitute the double angle formula of cosine function in denominator:
$\dfrac{{1 - \sin 2x}}{{\cos 2x}}$
We will now split the terms and it can be written as:
$\dfrac{1}{{\cos 2x}} - \dfrac{{\sin 2x}}{{\cos 2x}}$
We know that secant of an angle $x$ can be written as the reciprocal of cosine of the angle $x$ .
So by applying this, we can write
$\dfrac{1}{{\cos 2x}} = \sec 2x$
And we know that
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ ,
By comparing we have $\theta = 2x$
Therefore by applying this, we can write the expression as
$\sec 2x - \tan 2x$ .
Hence the required answer is $\sec 2x - \tan 2x$ .

Note:
We should note that after rationalising we have applied the difference of squares formula i.e.
${a^2} - {b^2} = (a + b)(a - b)$ .
So by comparing from the solution above, we have
$a = \cos x$
And
$b = \sin x$.
Therefore by applying this formula, it gives
$\left( {{{\cos }^2}x - {{\sin }^2}x} \right)$ .