Question

Which of the following trigonometric expressions is equivalent to csc(-112°)? csc(68°)

Answer 1

Answer:

Step-by-step explanation:

To find: $\csc \left ( -112^{\circ} \right )$

Solution:

Trigonometric expression is an expression consisting of trigonometric ratios like $\sin ,\tan ,\cos ,\csc ,\cot ,\sec$

Trigonometric ratios refers to the relation between the sides of right angled triangle and it's angles.

We know that angle $-112^{\circ}$ lies in the fourth quadrant in which $\csc$ is negative

So, $\csc \left ( -112^{\circ} \right )=-\csc \left ( 112^{\circ} \right )$

We can write $-\csc \left ( 112^{\circ} \right )=-\csc \left ( 180^{\circ}-68^{\circ} \right )$

We know that angle $180^{\circ}-68^{\circ}=112^{\circ}$ lies in second quadrant in which $\csc$ is positive

$-\csc \left ( 180^{\circ}-68^{\circ} \right )=-\csc68^{\circ}$

Therefore, we get

$\csc(-112^{\circ})=-\csc(112^{\circ})=-\csc \left ( 180^{\circ}-68^{\circ} \right )=-\csc68^{\circ}$

Answer 2

Answer:

csc(-112°)=-csc(68°)

Step-by-step explanation:

Since cosec(x) and sin(x) are reciprocals of each other,

cosec(-112°)=$\frac{1}{sin(-112^{\circ})}$

                  =$\frac{1}{-sin(112^{\circ})}$

                 =$\frac{1}{-sin(180^{\circ}-68^{\circ})}$     (since 112=180-68)

                 =$\frac{1}{-sin(68^{\circ})}$              (since sin(180-x)=sinx)

                 =$\frac{-1}{sin(68^{\circ})}$               (since $\frac{a}{-b}= \frac{-a}{b}$)

                 =-cosec(68°)