Question

How do you prove \[\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{1 - \cos x}}{{2 + \cos x}}\].

Answer 1

Hint: We solve the left hand side of the equation by using the identity \[{\sin ^2}x = 1 - {\cos ^2}x\]in the numerator and by solving the denominator by factoring it. Form factors for the quadratic equation in denominator and cancel possible factors from numerator and denominator.
* Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is \[(x - p)(x - q) = 0\]

Complete step by step solution:
We have to prove that \[\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{1 - \cos x}}{{2 + \cos x}}\]
Left hand side of the equation is \[\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}}\]
Substitute the value of \[{\sin ^2}x = 1 - {\cos ^2}x\]in the numerator of the fraction
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}}\]
Now we know that \[{a^2} - {b^2} = (a - b)(a + b)\], then we can write the numerator of the equation as \[1 - {\cos ^2}x = (1 - \cos x)(1 + \cos x)\]
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{(1 - \cos x)(1 + \cos x)}}{{{{\cos }^2}x + 3\cos x + 2}}\]
Now we factorize the denominator of the fraction. We can break \[3\cos x = \cos x + 2\cos x\]
 \[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{(1 - \cos x)(1 + \cos x)}}{{{{\cos }^2}x + \cos x + 2\cos x + 2}}\]
Take \[\cos x\]common from first two terms and 2 from last two terms in the denominator
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{(1 - \cos x)(1 + \cos x)}}{{\cos x(\cos x + 1) + 2(\cos x + 1)}}\]
Collect the common factors
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{(1 - \cos x)(1 + \cos x)}}{{(\cos x + 1)(\cos x + 2)}}\]
Cancel common factors from numerator and denominator
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{(1 - \cos x)}}{{(\cos x + 2)}}\]
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}} = \dfrac{{1 - \cos x}}{{2 + \cos x}}\]
This is equal to right hand side of the equation
So, left hand side of the equation is equal to right hand side of the equation
Hence proved

Note:
Many students make the mistake of cross multiplying the fractions on both sides of the equation in order to solve for the value of x which is wrong. Keep in mind we have to prove the left hand side of the equation is equal to the right hand side of the equation.
Alternate method:
We can also start by solving the right hand side of the equation. Multiply right hand side of the equation both numerator and denominator by \[1 + \cos x\]
\[ \Rightarrow \dfrac{{1 - \cos x}}{{2 + \cos x}} \times \dfrac{{1 + \cos x}}{{1 + \cos x}}\]
Use the property \[{a^2} - {b^2} = (a - b)(a + b)\]for the numerator
\[ \Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{2(1 + \cos x) + \cos x(1 + \cos x)}}\]
Substitute the value of \[{\sin ^2}x = 1 - {\cos ^2}x\]in the numerator of the fraction
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{2 + 2\cos x + \cos x + {{\cos }^2}x}}\]
Add the similar terms in the denominator of the fraction
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x + 3\cos x + 2}}\]
This is left hand side of the equation
So, left hand side of the equation is equal to right hand side of the equation
Hence proved