Question

Value of \[\cos ec( - {1410}^{\circ})\] is
A) \[\dfrac{1}{2}\]
B) \[ - \dfrac{1}{2}\]
C) \[\dfrac{{\sqrt 3 }}{2}\]
D) \[2\]

Answer 1

Hint: Trigonometric functions are the real functions which relate the angle of the right angled triangle to ratios of two side length. We have to take a round from the positive x axis, according to the angle and we know ${360}^{\circ}$ means 1 complete round.

Complete step by step solution:
Basic trigonometric functions:-
There are six basic trigonometric functions. Sine, Cosine, tangent, cotangent, Secant and cosecant. These are related to each other.
Let us discuss these functions from the basic concept of right angle triangle. Let \[\Delta ABC\]
Where \[\angle B = {90^{\circ}}\]
So the \[\sin \theta \] will be perpendicular / hypogynous
The \[\cos \theta \]will be Base / hypogynous
The \[\tan \theta \] will be perpendicular / Base
There are many formulas related to trigonometry functions.
In the given question it is based upon the \cos ec function. Let’s try to solve it.
\[\cos ec( - {1410}^{\circ})\] is the given function.
So it can be write it as
\[ = \cos ec\left( {1410 \times \dfrac{\pi }{{{{180}^{\circ}}}}} \right)\]
as from the identity \cos ec \[\cos ec( - x)\]
\[ = \cos ec x\]
So,
\[ = \cos ec\left( {1410 \times \dfrac{\pi }{{{{180}^{\circ}}}}} \right)\]
\[ = \cos ec\left( {141 \times \dfrac{\pi }{{{{18}^{\circ}}}}} \right)\]
Solving the above term as…
\[ = \cos ec\left( {\dfrac{{141\pi }}{{18}}} \right)\]…………………… \[(A)\]
Convert \[\dfrac{{141}}{{18}}\] to as lowest term like \[\dfrac{{141}}{{18}}\] dividing to a similar table\[3\]. we get
\[\dfrac{{141}}{{18}}\] = \[\dfrac{{47}}{6}\]
So, by dividing to similar table the get \[\dfrac{{47}}{6}\]
Now the equation \[(A)\] becomes…
\[ = - \cos ec\left( {\dfrac{{47}}{6}\pi } \right)\]
\[ = \cos ec\left( {\dfrac{{47}}{6}\pi } \right)\]
Now, diving \[\dfrac{{47}}{6}\] as the rate of \[Q\dfrac{R}{D}\], where \[R\] is the remainder = 5
\[Q\] is the Quotient = 7
\[D\] is the Diviser = 6
So, it will becomes \[7\dfrac{5}{6}\]
5 is the remainder, 6 will be the Diviser and 7 will be the Quotient.
\[7 + \dfrac{5}{6}\] or \[8 - \dfrac{1}{6}\]
we will take \[8 - \dfrac{1}{6}\] as it is even
Even numbers are the numbers which are divisible by \[2\]. The value of cosec x repeats after an internal of \[2\pi \]. So the equation becomes
\[ = - \cos ec\left( {8\pi - \dfrac{1}{6}\pi } \right)\]
\[ = - \cos ec\left( { - \dfrac{1}{6}\pi } \right)\]
Where the value of \[\pi \] is \[{{80}^{\circ}}\]
Substitute the value in above term
We get
\[ = - \cos ec\left( {\dfrac{{ - 1}}{6} \times {{180}^{\circ}}} \right)\]
Solving the above term
\[ = - \cos ec\left( {\dfrac{{ - 1}}{6} \times {{180}^{\circ}}} \right)\]
\[ = - \cos ec\left( { - {{30}^{\circ}}} \right)\]
as from identity rule \[\cos ec\left( { - x} \right)\]
- use $x$ So the equation becomes cosec \[{30^{\circ}}\] as from the formulas of trigonometry \[\cos ec \theta = \dfrac{1}{{\sin \theta }}\]
So the above equation can be written as
\[\cos ec{30^{\circ}}\dfrac{1}{{\sin {{30}^{\circ}}}}\]……….. \[(B)\]
as the value of \[\sin {30^{\circ}}\dfrac{1}{2}\]
Substituting the value of \[\sin {30^{\circ}} = \dfrac{1}{2}\]
In equation \[(B)\]we get
\[ = \dfrac{1}{{\sin {{30}^{\circ}}}}\]
\[ = \dfrac{1}{{\dfrac{1}{2}}}\]
\[ = \dfrac{2}{1}\]
\[ = 2\]
Hence the \[\cos ec\left( { - {{1410}^{\circ}}} \right)\] is 2.

Note:
The above questions are solved by using trigonometric functions and formulae. The trigonometric functions are widely used in all science that are related to Geometry. Such as navigations, Solid mechanics.