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How do you verify $\dfrac{{1 - \cos x}}{{1 + \cos x}} = {\left( {\csc x - \cot x} \right)^2}?$

seo-qna
Last updated date: 23rd Feb 2024
Total views: 339.6k
Views today: 3.39k
IVSAT 2024
Answer
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Hint:Multiply and divide L.H.S. (Left hand side) by the conjugate of its denominator. And then use some basic trigonometric identities like sine and cosine relation, sine and cosec relation, and cosine, sine and cot relation. After simplifying this all the left hand side will become equal to the right hand side (R.H.S.). Hence the problem is solved and verified.

Formula used:
$(a + b)(a - b) = {a^2} - {b^2}$
${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step answer:
In order to verify the given equation $\dfrac{{1 - \cos x}}{{1 + \cos x}} = {\left( {\csc x - \cot x} \right)^2}$ we need to prove \[{\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.,}}\] where they are left hand side and right hand side respectively. We will solve left hand side $\left( {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \right)$ to make it equal to the right hand side $\left( {{{\left( {\csc x - \cot x} \right)}^2}} \right)$
So let us start,
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$
Multiplying and dividing it with the conjugate of its denominator, we will get
$\dfrac{{1 - \cos x}}{{1 + \cos x}} \times \dfrac{{1 - \cos x}}{{1 - \cos x}} \\
\Rightarrow\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}} \\$
From the algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$ simplifying further,
\[\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}} \\
\Rightarrow\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {{1^2} - {{\cos }^2}x} \right)}} \\
\Rightarrow\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {1 - {{\cos }^2}x} \right)}} \\ \]
Using the trigonometric identity of sine and cosine that is the sum of square of sine and cosine equals to one, mathematically it can be written as
${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Simplifying further using this we will get,
\[\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{\left( {1 - {{\cos }^2}x} \right)}} \\
\Rightarrow\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{{{\sin }^2}x}} \\ \]
We can further write it as
\[\dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{{{\sin }^2}x}} \\
\Rightarrow{\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)^2} \\ \]
Now distributing the denominator over numerator using the distributive property of division,
\[{\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)^2} \\
\Rightarrow{\left( {\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}} \right)^2} \\ \]
From the relation between trigonometric functions we know that,
$\csc x = \dfrac{1}{{\sin x}}\;and\;\cot x = \dfrac{{\cos x}}{{\sin x}}$
Substituting these values above, we will get
\[{\left( {\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}} \right)^2} \\
\Rightarrow{\left( {\csc x - \cot x} \right)^2} \\
\therefore{\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \\ \]
Hence \[{\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}\] proved.

Note:You can also start to solve or proof this question from right hand side and end up to left hand side, you have to just decide which way is easier to you wither left hand side to right hand side or right hand side to left hand side.