Question

Value of $cosec( - {1110^ \circ })$
A) $\dfrac{2}{{\sqrt 3 }}$
B) $- \dfrac{2}{{\sqrt 3 }}$
C) $2$
D) $- 2$

Answer 1

Here we are asked to calculate the value of $cosec( - {1110^ \circ })$
We need to take the minus sign from the outside of the bracket and then we need to write the number $1110$ in the form of $90$, so the number $1110$ will be divided by $90$. Then we need to solve it and need to take reciprocal for the cosecant to obtain the desired answer.

Formula to be used:
Some trigonometric identities that will be used are given below:
$\cos ec( - \theta ) = - \cos ec(\theta )$
$\cos ec(n{360^ \circ } + \theta ) = \cos ec(\theta )$
$\cos ec(\theta ) = \dfrac{1}{{\sin (\theta )}}$

Complete answer:
We are given cosec$( - {1110^ \circ })$, and we are required to find its value.
To start with, use the identity given by
$\cos ec( - \theta ) = - \cos ec(\theta )$
To write the given trigonometric function as
$\cos ec( - {1110^ \circ }) = - \cos ec({1110^ \circ })$
Now, we need to factor the number $1110$ in the form of $90$, so the number $1110$ will be divided by $90$.
By dividing the number $1110$ by $90$, we found that the quotient is $12$ and the remainder is $30$.
So the number $1110$ can be written as (use: $Dividend = (Divisor \times Quotient) + Remainder$)
$1110 = (90 \times 12) + 30$
$- \cos ec({1110^ \circ }) = - \cos ec(12 \times {90^ \circ } + {30^ \circ })$
Since, we know that completing $4$rounds of ${90^ \circ }$ will complete ${360^ \circ }$, that is the angle will again come at ${0^ \circ }$.
$- \cos ec({1110^ \circ }) = - \cos ec(3(4 \times {90^ \circ }) + {30^ \circ })$
$- \cos ec({1110^ \circ }) = - \cos ec(3({360^ \circ }) + {30^ \circ })$
So, the given function will lie in the first quadrant which is completed by completing four revolutions ${360^ \circ }$.
Since we know that all the trigonometric functions are positive in the first quadrant. So the given function can be written as
(using the identity: $\cos ec(n \times {360^ \circ } + \theta ) = \cos ec(\theta )$)
$- \cos ec(3({360^ \circ }) + {30^ \circ }) = - \cos ec({30^ \circ })$
Now, we need to evaluate the value of $\cos ec({30^ \circ })$. Since, $\cos ec(\theta ) = \dfrac{1}{{\sin (\theta )}}$, we get
$- \cos ec({30^ \circ }) = \dfrac{{ - 1}}{{\sin ({{30}^ \circ })}}$
Since, the value of $\sin {30^ \circ }$ is $\dfrac{1}{2}$, substitute the value in the above trigonometric function,
$\dfrac{{ - 1}}{{\sin {{30}^ \circ }}} = \dfrac{{ - 1}}{{1/2}} = 2$
Hence, we get,
$\cos ec( - {1110^ \circ }) = - 2$
So, the value of $\cos ec( - {1110^ \circ })$ is (D)$- 2$.

Therefore, the correct option is D

Note: The Cartesian plane is divided into four quadrants, where all the trigonometric functions are positive in the first quadrant, only sine and cosine functions are positive in the second quadrant, only tangent and cotangent functions are positive in the third quadrant and the fourth quadrant, only secant and cosecant functions.