Evaluate : cos( 70 degree ) / sin ( 20 degree ) = cos ( 59 degree ) × cosec (31 degree ).
The Correct Answer and Explanation is:
To evaluate the given expression:cos(70∘)sin(20∘)=cos(59∘)×csc(31∘)\frac{\cos(70^\circ)}{\sin(20^\circ)} = \cos(59^\circ) \times \csc(31^\circ)sin(20∘)cos(70∘)=cos(59∘)×csc(31∘)
We will break down each part and verify if the two sides are equal.
Step 1: Simplify Left-Hand Side
First, we simplify the left-hand side, which is:cos(70∘)sin(20∘)\frac{\cos(70^\circ)}{\sin(20^\circ)}sin(20∘)cos(70∘)
Recall the trigonometric identity:cos(70∘)=sin(20∘)\cos(70^\circ) = \sin(20^\circ)cos(70∘)=sin(20∘)
This is because of the complementary angle identity, which states that:cos(θ)=sin(90∘−θ)\cos(\theta) = \sin(90^\circ – \theta)cos(θ)=sin(90∘−θ)
Thus, we have:cos(70∘)sin(20∘)=sin(20∘)sin(20∘)=1\frac{\cos(70^\circ)}{\sin(20^\circ)} = \frac{\sin(20^\circ)}{\sin(20^\circ)} = 1sin(20∘)cos(70∘)=sin(20∘)sin(20∘)=1
Step 2: Simplify Right-Hand Side
Now, simplify the right-hand side, which is:cos(59∘)×csc(31∘)\cos(59^\circ) \times \csc(31^\circ)cos(59∘)×csc(31∘)
Recall that:csc(θ)=1sin(θ)\csc(\theta) = \frac{1}{\sin(\theta)}csc(θ)=sin(θ)1
So, we have:cos(59∘)×csc(31∘)=cos(59∘)×1sin(31∘)\cos(59^\circ) \times \csc(31^\circ) = \cos(59^\circ) \times \frac{1}{\sin(31^\circ)}cos(59∘)×csc(31∘)=cos(59∘)×sin(31∘)1
Using the complementary angle identity again, we know:cos(59∘)=sin(31∘)\cos(59^\circ) = \sin(31^\circ)cos(59∘)=sin(31∘)
Thus:cos(59∘)×1sin(31∘)=sin(31∘)×1sin(31∘)=1\cos(59^\circ) \times \frac{1}{\sin(31^\circ)} = \sin(31^\circ) \times \frac{1}{\sin(31^\circ)} = 1cos(59∘)×sin(31∘)1=sin(31∘)×sin(31∘)1=1
Step 3: Conclusion
Both the left-hand side and right-hand side simplify to 1:cos(70∘)sin(20∘)=cos(59∘)×csc(31∘)=1\frac{\cos(70^\circ)}{\sin(20^\circ)} = \cos(59^\circ) \times \csc(31^\circ) = 1sin(20∘)cos(70∘)=cos(59∘)×csc(31∘)=1
Therefore, the given equation is true.
