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Question

Question: Determine the value of \(\sec ( - {45^\circ })\) \({\text{(A) }}\sqrt {\text{2}} \) \({\text{(B)...

Determine the value of sec(45)\sec ( - {45^\circ })
(A) 2{\text{(A) }}\sqrt {\text{2}}
(B) - 2{\text{(B) - }}\sqrt {\text{2}}
(C) 1{\text{(C) }}\sqrt 1
(D) - 1{\text{(D) - }}\sqrt 1

Explanation

Solution

First we must convert the negative angle to positive angle. We will first see where the specific angle lies in the angle graph and then use the appropriate formula to calculate the value.

Formula used: sec(θ)=1cos(θ)\sec (\theta ) = \dfrac{1}{{\cos (\theta )}}
cos(A+B)=cosAcosBSinASinB\cos (A + B) = \cos A\cos B - \operatorname{Sin} A\operatorname{Sin} B

Complete step-by-step answer:
We know that sec(θ)=1cos(θ)\sec (\theta ) = \dfrac{1}{{\cos (\theta )}}
So we have to put the value of θ\theta
\Rightarrow sec(45)=1cos(45)\sec ( - 45^\circ ) = \dfrac{1}{{\cos ( - 45^\circ )}}

From the graph we can see that the angle 45 - {45^ \circ } corresponds to angle 315315^\circ .
Therefore, angle 315315^\circ can be written as a combination of 270+45270^\circ + 45^\circ
Therefore, the question can be rewritten as:
\Rightarrow sec(315)=1cos(315)\sec (315^\circ ) = \dfrac{1}{{\cos (315^\circ )}}
This could further be written as:
\Rightarrow sec(270+45)=1cos(270+45).(1)\sec (270^\circ + 45^\circ ) = \dfrac{1}{{\cos (270^\circ + 45^\circ )}} ………………. (1)
Now let’s consider cos(270+45)\cos (270^\circ + 45^\circ )
We know that cos(A+B)=cosAcosBSinASinB\cos (A + B) = \cos A\cos B - \operatorname{Sin} A\operatorname{Sin} B
Therefore, the problem could be expanded as:
\Rightarrow cos(270+45)=cos(270)cos(45)sin(270)sin(45)\cos (270^\circ + 45^\circ ) = \cos (270^\circ )\cos (45^\circ ) - \sin (270^\circ )\sin (45^\circ )
Now from the trigonometric table that:
cos(270)=0\cos (270^\circ ) = 0,
cos(45)=12\cos (45^\circ ) = \dfrac{1}{{\sqrt 2 }},
sin(270)=1\sin (270^\circ ) = - 1,
sin(45)=12\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }}.
On substituting the values, we get:
\Rightarrow cos(270+45)=0×12(1)×12\cos (270^\circ + 45^\circ ) = 0 \times \dfrac{1}{{\sqrt 2 }} - ( - 1) \times \dfrac{1}{{\sqrt 2 }}
On simplifying we get:
\Rightarrow cos(270+45)=12.................(2)\cos (270^\circ + 45^\circ ) = \dfrac{1}{{\sqrt 2 }}.................\left( 2 \right)
Now on substituting the value (2)\left( 2 \right) in equation (1)(1)we get:
\Rightarrow sec(270+45)=112\sec (270^\circ + 45^\circ ) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}
On simplifying we get:
\Rightarrow sec(270+45)=2\sec (270^\circ + 45^\circ ) = \sqrt 2
Hence sec(45)=2\sec ( - {45^\circ }) = \sqrt 2 is the required answer.

\therefore The correct answer is option (A)(A).

Note: A negative angle is an angle which is traced clockwise from 00^\circ , therefore at that same point when traced anti-clockwise, would be another angle which is the same angle but positive.
In these types of questions, the various trigonometric formulas should be remembered such as the angle formulas and the sum and difference identities should also be remembered.
A graph should be made to visualize the angle on which the calculation is to be made.
There directly won’t be sum and difference identities for sec\theta, cosec\thetaor cot\theta{\text{sec\theta , cosec\theta or cot\theta }} so therefore they should be converted into their respective identities and the formulas should be then applied. Thereafter it should not be forgotten to convert the solution back to the original format because not converting it would result in an incomplete answer.