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

Value of csc⁡(−1410∘)\csc ( - 1410^\circ ) is:
A.12\dfrac{1}{2}
B. −12 - \dfrac{1}{2}
C. 32\dfrac{{\sqrt 3 }}{2}
D. 22

Explanation

Solution

To solve the above question, we will simplify the negative sign present in the angle with the help of sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x formula, and from the reference of graph we will ignore multiples of 360∘360^\circ or 2π2\pi . As the value of csc⁡\csc repeats after 360∘360^\circ or 2π2\pi intervals.
Formula used:
sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x

Complete step-by-step answer:
We want to find the value ofcsc⁡(−1410∘)\csc ( - 1410^\circ )
First a fall, we know that sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x, but as per the question we want csc⁡\csc .
So, we will take reciprocal on the both side of formula.
1sin⁡(−x)=−1sin⁡x......(1)\dfrac{1}{{\sin ( - x)}} = - \dfrac{1}{{\sin x}}......(1)
We know that 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta put this value in equation (1)(1)
Thus, we can write csc⁡(−x)=−csc⁡(x)\csc ( - x) = - \csc (x)
Now, we can write our question csc⁡(1410∘)=−csc⁡(1410∘)\csc (1410^\circ ) = - \csc (1410^\circ )
Here, we will convert 1410 degree into radians by multiply π180∘\dfrac{\pi }{{180^\circ }} with angle
=−csc⁡(1410∘×π180∘)= - \csc \left( {1410^\circ \times \dfrac{\pi }{{180^\circ }}} \right)
After simplification we get −csc⁡(47π6) - \csc \left( {\dfrac{{47\pi }}{6}} \right)
We can write above equation also like below equations
=−csc⁡(7π+5π6)= - \csc \left( {7\pi + \dfrac{{5\pi }}{6}} \right) and −csc⁡(8π−π6) - \csc \left( {8\pi - \dfrac{\pi }{6}} \right)
We will continue our solution with 8π−π68\pi - \dfrac{\pi }{6} as it is an even number.
As per the csc⁡\csc graph, we can say that the value of csc⁡\csc repeats after the 2π2\pi intervals
Thus, we will consider −csc⁡(8π−π6)=−csc⁡(−π6) - \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⁡(−π6)=csc⁡(π6) - \csc \left( { - \dfrac{\pi }{6}} \right) = \csc \left( {\dfrac{\pi }{6}} \right)
Here we will convert radians into degree by putting value of π=180∘\pi = 180^\circ
=csc⁡(180∘6)=csc⁡30∘= \csc \left( {\dfrac{{180^\circ }}{6}} \right) = \csc 30^\circ
And csc⁡30∘=1sin⁡30∘\csc 30^\circ = \dfrac{1}{{\sin 30^\circ }}
=112= \dfrac{1}{{\dfrac{1}{2}}}(Putsin⁡30∘=12\sin 30^\circ = \dfrac{1}{2})
=2= 2
Hence option D is the right option.

Note: Second method to solve above question.
csc⁡(−1410∘)\csc ( - 1410^\circ )
We know formulasin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x
We will take reciprocal on the both side of formula, as we want csccsc
1sin⁡(−x)=−1sin⁡x......(1)\dfrac{1}{{\sin ( - x)}} = - \dfrac{1}{{\sin x}}......(1)
We know that 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta put this value in equation (1)(1)
csc⁡(−x)=−csc⁡(x)\csc ( - x) = - \csc (x)
Now, we can write our question csc⁡(−1410∘)=−csc⁡(1410∘)\csc ( - 1410^\circ ) = - \csc (1410^\circ )
We will expand the angle of the given question, as we know that the value of csc⁡\csc repeats after the 360∘360^\circ intervals. So, we will ignore 360∘360^\circ in the above equation.
=−csc⁡(360∘×4−30)=−csc⁡(−30∘)= - \csc (360^\circ \times 4 - 30) = - \csc ( - 30^\circ )
Again as per above explanation we can write −csc⁡(−30∘)=−(−csc⁡(30∘)) - \csc ( - 30^\circ ) = - ( - \csc (30^\circ ))
Product of two negative will be positive as below
csc⁡30∘=1sin⁡30∘\csc 30^\circ = \dfrac{1}{{\sin 30^\circ }}
=112= \dfrac{1}{{\dfrac{1}{2}}}(Putsin⁡30∘=12\sin 30^\circ = \dfrac{1}{2})
=2= 2
Hence option D is the right option