Question

Value of \[\csc ( - 1410^\circ )\] is:
A.$\dfrac{1}{2}$
B. $ - \dfrac{1}{2}$
C. $\dfrac{{\sqrt 3 }}{2}$
D. $2$

Answer 1

Hint: To solve the above question, we will simplify the negative sign present in the angle with the help of $\sin ( - x) = - \sin x$ formula, and from the reference of graph we will ignore multiples of $360^\circ $ or $2\pi $ . As the value of $\csc $ repeats after $360^\circ $ or $2\pi $ intervals.
Formula used:
$\sin ( - x) = - \sin x$

Complete step-by-step answer:
We want to find the value of\[\csc ( - 1410^\circ )\]
First a fall, we know that $\sin ( - x) = - \sin x$, but as per the question we want $\csc $.
So, we will take reciprocal on the both side of formula.
$\dfrac{1}{{\sin ( - x)}} = - \dfrac{1}{{\sin x}}......(1)$
We know that $\dfrac{1}{{\sin \theta }} = \csc \theta $ put this value in equation $(1)$
Thus, we can write $\csc ( - x) = - \csc (x)$
Now, we can write our question \[\csc (1410^\circ ) = - \csc (1410^\circ )\]
Here, we will convert 1410 degree into radians by multiply $\dfrac{\pi }{{180^\circ }}$ with angle
\[ = - \csc \left( {1410^\circ \times \dfrac{\pi }{{180^\circ }}} \right)\]
After simplification we get \[ - \csc \left( {\dfrac{{47\pi }}{6}} \right)\]
We can write above equation also like below equations
$ = - \csc \left( {7\pi + \dfrac{{5\pi }}{6}} \right)$ and $ - \csc \left( {8\pi - \dfrac{\pi }{6}} \right)$
We will continue our solution with $8\pi - \dfrac{\pi }{6}$ as it is an even number.
As per the $\csc $graph, we can say that the value of $\csc $ repeats after the $2\pi $intervals
Thus, we will consider $ - \csc \left( {8\pi - \dfrac{\pi }{6}} \right) = - \csc \left( { - \dfrac{\pi }{6}} \right)$
Again as per above explanation we can write and product of two negative will be positive as below:$ - \csc \left( { - \dfrac{\pi }{6}} \right) = \csc \left( {\dfrac{\pi }{6}} \right)$
Here we will convert radians into degree by putting value of $\pi = 180^\circ $
$ = \csc \left( {\dfrac{{180^\circ }}{6}} \right) = \csc 30^\circ $
And $\csc 30^\circ = \dfrac{1}{{\sin 30^\circ }}$
$ = \dfrac{1}{{\dfrac{1}{2}}}$(Put$\sin 30^\circ = \dfrac{1}{2}$)
$ = 2$
Hence option D is the right option.

Note: Second method to solve above question.
\[\csc ( - 1410^\circ )\]
We know formula$\sin ( - x) = - \sin x$
We will take reciprocal on the both side of formula, as we want $csc$
$\dfrac{1}{{\sin ( - x)}} = - \dfrac{1}{{\sin x}}......(1)$
We know that $\dfrac{1}{{\sin \theta }} = \csc \theta $put this value in equation $(1)$
$\csc ( - x) = - \csc (x)$
Now, we can write our question \[\csc ( - 1410^\circ ) = - \csc (1410^\circ )\]
We will expand the angle of the given question, as we know that the value of $\csc $ repeats after the $360^\circ $ intervals. So, we will ignore $360^\circ $ in the above equation.
$ = - \csc (360^\circ \times 4 - 30) = - \csc ( - 30^\circ )$
Again as per above explanation we can write $ - \csc ( - 30^\circ ) = - ( - \csc (30^\circ ))$
Product of two negative will be positive as below
$\csc 30^\circ = \dfrac{1}{{\sin 30^\circ }}$
$ = \dfrac{1}{{\dfrac{1}{2}}}$(Put$\sin 30^\circ = \dfrac{1}{2}$)
$ = 2$
Hence option D is the right option