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How do you simplify the identity $\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x$?

Last updated date: 25th Jul 2024
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Hint: In the above question, the concept is based on the concept of trigonometry. The main approach towards solving this equation is applying trigonometric identity so that it further becomes easy to simplify by cancelling the common terms in numerator and denominator and prove the terms on the right-hand side.

Complete step by step solution:
Trigonometric function defines the function of the angle between the two sides. It gives us the relation between the angles sides of the right-angle triangle and angles. These trigonometric functions also have different trigonometric identities which help in simplifying equations.
Given is the equation we need to prove
$\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x$
So, on solving the left-hand side
\[\dfrac{{{{\sin }^2}x}}{{1 - \cos x}}\]
So now by applying the trigonometric identity which is given below,
\[{\sin ^2}x + {\cos ^2}x = 1\]
Therefore, by shifting the term of cosine function on the other side.
\[{\sin ^2}x = 1 - {\cos ^2}x\]
\[\dfrac{{1 - {{\cos }^2}x}}{{1 - \cos x}}\]
Now by applying the formula of difference of two terms in the numerator we write in the below given form:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Now by applying the above formula we get,
\[\dfrac{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}{{1 - \cos x}}\]
By cancelling out common terms in the numerator and denominator we get,
\[\left( {1 + \cos x} \right)\]
Therefore, the above term we get is equal to the terms on the right-hand side in the above given question. Hence the above equation is proved.

Note: An important thing to note is that the trigonometric identity is actually a Pythagorean identity. Pythagora's trigonometric identity is also called with another name as the Pythagorean identity.It is an identity expressing the Pythagorean theorem in terms of trigonometric functions. It explains the basic relation between sine and cosine function.