Respuesta :
Answer:
csc(-112°)=-csc(68°)
Step-by-step explanation:
Since cosec(x) and sin(x) are reciprocals of each other,
cosec(-112°)=[tex]\frac{1}{sin(-112^{\circ})}[/tex]
=[tex]\frac{1}{-sin(112^{\circ})}[/tex]
=[tex]\frac{1}{-sin(180^{\circ}-68^{\circ})}[/tex] (since 112=180-68)
=[tex]\frac{1}{-sin(68^{\circ})}[/tex] (since sin(180-x)=sinx)
=[tex]\frac{-1}{sin(68^{\circ})}[/tex] (since [tex]\frac{a}{-b}= \frac{-a}{b}[/tex])
=-cosec(68°)
Answer:
Step-by-step explanation:
To find: [tex]\csc \left ( -112^{\circ} \right )[/tex]
Solution:
Trigonometric expression is an expression consisting of trigonometric ratios like [tex]\sin ,\tan ,\cos ,\csc ,\cot ,\sec[/tex]
Trigonometric ratios refers to the relation between the sides of right angled triangle and it's angles.
We know that angle [tex]-112^{\circ}[/tex] lies in the fourth quadrant in which [tex]\csc[/tex] is negative
So, [tex]\csc \left ( -112^{\circ} \right )=-\csc \left ( 112^{\circ} \right )[/tex]
We can write [tex]-\csc \left ( 112^{\circ} \right )=-\csc \left ( 180^{\circ}-68^{\circ} \right )[/tex]
We know that angle [tex]180^{\circ}-68^{\circ}=112^{\circ}[/tex] lies in second quadrant in which [tex]\csc[/tex] is positive
[tex]-\csc \left ( 180^{\circ}-68^{\circ} \right )=-\csc68^{\circ}[/tex]
Therefore, we get
[tex]\csc(-112^{\circ})=-\csc(112^{\circ})=-\csc \left ( 180^{\circ}-68^{\circ} \right )=-\csc68^{\circ}[/tex]
