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Question: How do you verify the identity \[\dfrac{{\sec \theta - 1}}{{1 - \cos \theta }} = \sec \theta \] ?...

How do you verify the identity sec⁡θ−11−cos⁡θ=sec⁡θ\dfrac{{\sec \theta - 1}}{{1 - \cos \theta }} = \sec \theta ?

Explanation

Solution

Hint : This problem is related to the trigonometric identities. Here we will just use two basic identities and that are already given in the question only. That is the cos function and sec function. These are the reciprocals of each other. We will start with LHS to prove RHS.

Complete step-by-step answer :
Given that,
sec⁡θ−11−cos⁡θ\dfrac{{\sec \theta - 1}}{{1 - \cos \theta }}
We know that sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }}
Now substitute this value in the numerator,
=1cos⁡θ−11−cos⁡θ= \dfrac{{\dfrac{1}{{\cos \theta }} - 1}}{{1 - \cos \theta }}
Now taking LCM on the numerator,
=1−cos⁡θcos⁡θ1−cos⁡θ= \dfrac{{\dfrac{{1 - \cos \theta }}{{\cos \theta }}}}{{1 - \cos \theta }}
Taking the cos term in denominator like abc=abc\dfrac{{\dfrac{a}{b}}}{c} = \dfrac{a}{{bc}}
=1−cos⁡θ(1−cos⁡θ)cos⁡θ= \dfrac{{1 - \cos \theta }}{{\left( {1 - \cos \theta } \right)\cos \theta }}
Cancelling the same terms from the numerator and denominator,
=1cos⁡θ= \dfrac{1}{{\cos \theta }}
We know that sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }}
=sec⁡θ= \sec \theta
And this is nothing but,
=RHS= RHS
Hence proved.

Note : Note that the problem given is very simple. We just need to use two trigonometric functions which are coincidently reciprocals of each other. But note that if there are any other functions then also we will use identities of trigonometry but those which will be leading towards the solution. Never complicate these types of problems and always write each and every step.