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Question: The value of \[\sec \theta \] is equals to 1\. \[\dfrac{1}{{\sqrt {1 - {{\cos }^2}\theta } }}\] ...

The value of sec⁡θ\sec \theta is equals to
1. 11−cos⁡2θ\dfrac{1}{{\sqrt {1 - {{\cos }^2}\theta } }}
2. 1+cot⁡2θcot⁡θ\dfrac{{\sqrt {1 + {{\cot }^2}\theta } }}{{\cot \theta }}
3. cot⁡θ1+cot⁡2θ\dfrac{{\cot \theta }}{{\sqrt {1 + {{\cot }^2}\theta } }}
4. cosec⁡2θ−1cosec⁡θ\dfrac{{\sqrt {{{\operatorname{cosec} }^2}\theta - 1} }}{{\operatorname{cosec} \theta }}

Explanation

Solution

The term sec⁡θ\sec \theta is a trigonometric ratio . It is also known as the reciprocal of cos⁡θ\cos \theta . In the given question we must simplify the given options to sec⁡θ\sec \theta . So , we will solve the given options one by one and we will use three basic identities of trigonometry which are sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta and 1+cot⁡2θ=cosec⁡2θ1 + {\cot ^2}\theta = {\operatorname{cosec} ^2}\theta accordingly .

Complete step-by-step solution:
Solving option (1) we get,
=11−cos⁡2θ= \dfrac{1}{{\sqrt {1 - {{\cos }^2}\theta } }}
Using the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we get ,
=1sin⁡2θ= \dfrac{1}{{\sqrt {{{\sin }^2}\theta } }}
Solving the square root we get ,
=1sin⁡θ= \dfrac{1}{{\sin \theta }}
Taking the reciprocal of sin⁡θ\sin \theta we get ,
=cosec⁡θ= \operatorname{cosec} \theta .
Therefore , option (1) is not the correct answer .
Now , solving option (2) we get ,
=1+cot⁡2θcot⁡θ= \dfrac{{\sqrt {1 + {{\cot }^2}\theta } }}{{\cot \theta }}
Now using the identity 1+cot⁡2θ=cosec⁡2θ1 + {\cot ^2}\theta = {\operatorname{cosec} ^2}\theta we get ,
=cosec⁡2θcot⁡θ= \dfrac{{\sqrt {{{\operatorname{cosec} }^2}\theta } }}{{\cot \theta }}
Now solving the square root we get ,
=cosec⁡θcot⁡θ= \dfrac{{\operatorname{cosec} \theta }}{{\cot \theta }}
Now simplifying the ratios we get ,
=1sin⁡θcos⁡θsin⁡θ= \dfrac{{\dfrac{1}{{\sin \theta }}}}{{\dfrac{{\cos \theta }}{{\sin \theta }}}}
On solving we get ,
=1cos⁡θ= \dfrac{1}{{\cos \theta }}
Taking the reciprocal of cos⁡θ\cos \theta we get ,
=sec⁡θ= \sec \theta
Therefore , option (2) is the correct answer .
Now we will check other options too .
Now solving option (3) we get ,
=cot⁡θ1+cot⁡2θ= \dfrac{{\cot \theta }}{{\sqrt {1 + {{\cot }^2}\theta } }}
Now using the identity 1+cot⁡2θ=cosec⁡2θ1 + {\cot ^2}\theta = {\operatorname{cosec} ^2}\theta we get ,
=cot⁡θcosec⁡2θ= \dfrac{{\cot \theta }}{{\sqrt {{{\operatorname{cosec} }^2}\theta } }}
On solving we get ,
=cot⁡θcosec⁡θ= \dfrac{{\cot \theta }}{{\operatorname{cosec} \theta }}
On simplifying we get ,
=cos⁡θsin⁡θ1sin⁡θ= \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{\dfrac{1}{{\sin \theta }}}}
On solving we get ,
=cos⁡θ= \cos \theta
Therefore , option (3) is the wrong answer .
Now we will solve option (4) we get ,
=cosec⁡2θ−1cos⁡ecθ= \dfrac{{\sqrt {{{\operatorname{cosec} }^2}\theta - 1} }}{{\cos ec\theta }}
Now using the identity 1+cot⁡2θ=cosec⁡2θ1 + {\cot ^2}\theta = {\operatorname{cosec} ^2}\theta we get ,
=cot⁡2θcosec⁡θ= \dfrac{{\sqrt {{{\cot }^2}\theta } }}{{\operatorname{cosec} \theta }}
On solving we get ,
=cot⁡θcosec⁡θ= \dfrac{{\cot \theta }}{{\operatorname{cosec} \theta }}
On simplifying we get ,
=cos⁡θsin⁡θ1sin⁡θ= \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{\dfrac{1}{{\sin \theta }}}}
On solving we get ,
=cos⁡θ= \cos \theta
Therefore , option (4) is also the wrong answer.

Note: In these questions , which are related to the trigonometric ratio the basic step is that you should know about the three identities of trigonometric ratios . Try to solve the option rather than the question . There are also some questions where multiple answers are also there, so you should solve every given option.