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Question

Question: How do you find the value of\[sec\left( {315} \right)\] ?...

How do you find the value ofsec(315)sec\left( {315} \right) ?

Explanation

Solution

Hint : While solving this particular question we must convert the given trigonometric function into its corresponding simpler form that is sec(315)=1cos(315)\sec (315) = \dfrac{1}{{\cos (315)}} that manipulate 315=36045{315^ \circ } = {360^ \circ } - {45^ \circ } , then apply the even function identity , at the end you will get your desired result . Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.

Complete step-by-step answer :
We have to find the value of given trigonometric ratio that is sec(315)sec\left( {315} \right) ,
We already know the relationship between secant and cosine that is ,
secx=1cosx\sec x = \dfrac{1}{{\cos x}}
Therefore, we can write the given expression as ,
sec(315)=1cos(315)\sec (315) = \dfrac{1}{{\cos (315)}}
Here 315315 must be in the degree form, therefore , we can write this as 315=36045{315^ \circ } = {360^ \circ } - {45^ \circ } , and since 360{360^ \circ } is a complete loop ,
And we know that cos(360x)=cos(x)\cos ({360^ \circ } - x) = \cos ( - x) we will get the following result,
sec(315)=1cos(45)\sec (315) = \dfrac{1}{{\cos ( - 45)}}
Since we also know that the cosine is an even function, cos(x)=cos(x)cos( - x) = cos(x) , so
sec(315)=1cos(45)\sec (315) = \dfrac{1}{{\cos (45)}}
45{45^ \circ } is a special angle and we know that the cosine of 45{45^ \circ } is 12\dfrac{1}{{\sqrt 2 }} ,
Therefore, we will get the required result ,
sec(315)=112=2\sec (315) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} = \sqrt 2
Hence we get the solution.
So, the correct answer is “2\sqrt 2”.

Note : By using the basic trigonometric identity given below we can simplify and can find the value of the above expression that is sec(315)sec\left( {315} \right) . In order to solve and simplify the given expression we
have to use the identities and express our given expression in the simplest form and thereby solve
it. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.