Question

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

Answer 1

Hint: This question is related to the trigonometry, and we have to prove that the left hand side is equal to the right hand side of the expression, and this question can be solved by using trigonometric identities i.e.,\[\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}\] and\[\csc x = \dfrac{1}{{\sin x}}\], and now by simplifying the expression using algebraic identity \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], and by further simplifying we will get the result which is on the right hand.

Complete step by step solution:
Given \[{\left( {\csc x + \cot x} \right)^2} = \dfrac{{1 + \cos x}}{{1 - \cos x}}\],
Now we have to prove that left hand side is equal to the right hand side of the equation, now take the expression on the left hand side i.e.,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2}\],
Now using the trigonometric identities i.e., \[\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}\] and\[\csc x = \dfrac{1}{{\sin x}}\],
Now substituting the identities in the left hand side of the given expression we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = {\left( {\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}} \right)^2}\],
Now simplifying we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = {\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right)^2}\],
Now separating the numerator and denominator we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = \dfrac{{{{\left( {1 + \cos x} \right)}^2}}}{{{{\sin }^2}x}}\],
Now using the identity \[{\sin ^2}x = 1 - {\cos ^2}x\], we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = \dfrac{{{{\left( {1 + \cos x} \right)}^2}}}{{1 - {{\cos }^2}x}}\],
Now using the algebraic identity, \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\]in the denominator we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = \dfrac{{{{\left( {1 + \cos x} \right)}^2}}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}\],
Now eliminating the like terms we get,
\[ \Rightarrow {\left( {\csc x + \cot x} \right)^2} = \dfrac{{1 + \cos x}}{{1 - \cos x}}\],
Which is equal to the right hand side of the equation,
Hence proved.
Final Answer:
\[\therefore \]By using identities we proved that expression \[{\left( {\csc x + \cot x} \right)^2}\]is equal to \[\dfrac{{1 + \cos x}}{{1 - \cos x}}\].

Note:
An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\],
\[\sec x = \dfrac{1}{{\cos x}}\],
\[\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}\] ,
\[{\sin ^2}x = 1 - {\cos ^2}x\],
\[{\cos ^2}x + {\sin ^2}x = 1\],
\[{\sec ^2}x - {\tan ^2}x = 1\],
\[{\csc ^2}x = 1 + {\cot ^2}x\].
\[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\],
\[{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1\],
\[\sin 2x = 2\sin x\cos x\],
\[2{\cos ^2}x = 1 + \cos 2x\],
\[\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\].