If \[a = \sec {2^ \circ }\] and \[b = \sec 2\], then
A) \[a = b\]
B) \[a < b\]
C) \[b < a\]
D) \[2a = b\]
Answer
590.4k+ views
Hint: In the given question, we have to apply trigonometric identity to arrive at the solution. Since the question is given in degrees, we can solve it by converting it into radians. We can use the trigonometric ratio table to find out the required value.
Complete step by step solution:
Let us understand with the help of diagram:
The diagram shows four quadrants: I, II, III and IV along-with the degrees and their radians at the end of the line segment. All of the even multiples are the quadrantal angles and are reduced, just like any other fraction.
Now \[a = \sec {2^ \circ }\] is a positive value lying in quadrant I. Therefore, we can conclude that \[a > 0\] i.e. it is a positive value.
Since \[{180^ \circ } = \pi \], we can conclude that \[{1^ \circ } = \dfrac{\pi }{{180}}\]
Therefore, multiplying by \[2\] on both the sides, we will get,
\[{2^ \circ } = \dfrac{\pi }{{180}}(2) = \dfrac{\pi }{{90}}\]
Hence, we can conclude that:
\[a = \sec {2^ \circ } = \sec \dfrac{\pi }{{90}}\]
Now we are given that \[b = \sec 2\]
We know that \[\dfrac{\pi }{{90}} < 2\] since the value of \[\pi \] is approximate \[3.14\].
So, we can say that:
\[\sec {2^ \circ } > \sec 2\]
\[ \Rightarrow b < a\]
Hence, Option (C) \[b < a\] is the correct answer.
Note:
1) We can solve alternatively as follows:
Taking value of \[\pi \] as \[3.14\], we can calculate that-
\[{90^ \circ } = \dfrac{\pi }{2} = \dfrac{{3.14}}{2} = 1.57\]and
\[{180^ \circ } = \pi = 3.14\]
Now we know that \[2\]lies between the above two values i.e.
\[1.57 \leqslant 2 \leqslant 3.14\]
This means that the value of \[\sec 2\]lies between \[{90^ \circ }\] and \[{180^ \circ }\] i.e. in Quadrant II where \[\sec < 0\].
Hence, we can say that \[b < 0\].
Since \[\sec {2^ \circ }\] is positive and lying in Quadrant I, we can conclude that:
\[ \Rightarrow b < a\]
2) Followings points should be remembered in case of quadrants:
In Quadrant I: All trigonometric functions are positive.
In Quadrant II: Only \[\sin \] and \[\cos ec\] trigonometric functions are positive.
In Quadrant III: Only \[\tan \] and \[\cot \] trigonometric functions are positive.
In Quadrant IV: Only \[\sec \] and \[\cos \] trigonometric functions are positive.
Complete step by step solution:
Let us understand with the help of diagram:
The diagram shows four quadrants: I, II, III and IV along-with the degrees and their radians at the end of the line segment. All of the even multiples are the quadrantal angles and are reduced, just like any other fraction.
Now \[a = \sec {2^ \circ }\] is a positive value lying in quadrant I. Therefore, we can conclude that \[a > 0\] i.e. it is a positive value.
Since \[{180^ \circ } = \pi \], we can conclude that \[{1^ \circ } = \dfrac{\pi }{{180}}\]
Therefore, multiplying by \[2\] on both the sides, we will get,
\[{2^ \circ } = \dfrac{\pi }{{180}}(2) = \dfrac{\pi }{{90}}\]
Hence, we can conclude that:
\[a = \sec {2^ \circ } = \sec \dfrac{\pi }{{90}}\]
Now we are given that \[b = \sec 2\]
We know that \[\dfrac{\pi }{{90}} < 2\] since the value of \[\pi \] is approximate \[3.14\].
So, we can say that:
\[\sec {2^ \circ } > \sec 2\]
\[ \Rightarrow b < a\]
Hence, Option (C) \[b < a\] is the correct answer.
Note:
1) We can solve alternatively as follows:
Taking value of \[\pi \] as \[3.14\], we can calculate that-
\[{90^ \circ } = \dfrac{\pi }{2} = \dfrac{{3.14}}{2} = 1.57\]and
\[{180^ \circ } = \pi = 3.14\]
Now we know that \[2\]lies between the above two values i.e.
\[1.57 \leqslant 2 \leqslant 3.14\]
This means that the value of \[\sec 2\]lies between \[{90^ \circ }\] and \[{180^ \circ }\] i.e. in Quadrant II where \[\sec < 0\].
Hence, we can say that \[b < 0\].
Since \[\sec {2^ \circ }\] is positive and lying in Quadrant I, we can conclude that:
\[ \Rightarrow b < a\]
2) Followings points should be remembered in case of quadrants:
In Quadrant I: All trigonometric functions are positive.
In Quadrant II: Only \[\sin \] and \[\cos ec\] trigonometric functions are positive.
In Quadrant III: Only \[\tan \] and \[\cot \] trigonometric functions are positive.
In Quadrant IV: Only \[\sec \] and \[\cos \] trigonometric functions are positive.
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