
How do you find the value of\[sec\left( {315} \right)\] ?
Answer
525.6k+ views
Hint: While solving this particular question we must convert the given trigonometric function into its corresponding simpler form that is $ \sec (315) = \dfrac{1}{{\cos (315)}} $ that manipulate $ {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 \[sec\left( {315} \right)\] ,
We already know the relationship between secant and cosine that is ,
$ \sec x = \dfrac{1}{{\cos x}} $
Therefore, we can write the given expression as ,
$ \sec (315) = \dfrac{1}{{\cos (315)}} $
Here $ 315 $ must be in the degree form, therefore , we can write this as $ {315^ \circ } = {360^ \circ } - {45^ \circ } $ , and since $ {360^ \circ } $ is a complete loop ,
And we know that $ \cos ({360^ \circ } - x) = \cos ( - x) $ we will get the following result,
$ \sec (315) = \dfrac{1}{{\cos ( - 45)}} $
Since we also know that the cosine is an even function, \[cos( - x) = cos(x)\] , so
$ \sec (315) = \dfrac{1}{{\cos (45)}} $
$ {45^ \circ } $ is a special angle and we know that the cosine of $ {45^ \circ } $ is $ \dfrac{1}{{\sqrt 2 }} $ ,
Therefore, we will get the required result ,
$ \sec (315) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} = \sqrt 2 $
Hence we get the solution.
So, the correct answer is “$ \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 \[sec\left( {315} \right)\] . 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.
Complete step-by-step answer:
We have to find the value of given trigonometric ratio that is \[sec\left( {315} \right)\] ,
We already know the relationship between secant and cosine that is ,
$ \sec x = \dfrac{1}{{\cos x}} $
Therefore, we can write the given expression as ,
$ \sec (315) = \dfrac{1}{{\cos (315)}} $
Here $ 315 $ must be in the degree form, therefore , we can write this as $ {315^ \circ } = {360^ \circ } - {45^ \circ } $ , and since $ {360^ \circ } $ is a complete loop ,
And we know that $ \cos ({360^ \circ } - x) = \cos ( - x) $ we will get the following result,
$ \sec (315) = \dfrac{1}{{\cos ( - 45)}} $
Since we also know that the cosine is an even function, \[cos( - x) = cos(x)\] , so
$ \sec (315) = \dfrac{1}{{\cos (45)}} $
$ {45^ \circ } $ is a special angle and we know that the cosine of $ {45^ \circ } $ is $ \dfrac{1}{{\sqrt 2 }} $ ,
Therefore, we will get the required result ,
$ \sec (315) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} = \sqrt 2 $
Hence we get the solution.
So, the correct answer is “$ \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 \[sec\left( {315} \right)\] . 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

