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Math  /  Trigonometry

QuestionSimplify the trigonometric expression. 1+sin⁡(y)1+csc⁡(y)\frac{1+\sin (y)}{1+\csc (y)}

Studdy Solution

STEP 1

1. We are asked to simplify the trigonometric expression.
2. The expression involves sine and cosecant functions.
3. Simplification may involve rewriting the trigonometric functions in terms of sine.

STEP 2

1. Rewrite the cosecant function in terms of sine.
2. Simplify the expression by combining terms.

STEP 3

Rewrite the cosecant function in terms of sine. Recall that:
csc⁡(y)=1sin⁡(y) \csc(y) = \frac{1}{\sin(y)}
Substitute this into the expression:
1+sin⁡(y)1+1sin⁡(y) \frac{1+\sin(y)}{1+\frac{1}{\sin(y)}}

STEP 4

Simplify the denominator by finding a common denominator:
1+1sin⁡(y)=sin⁡(y)sin⁡(y)+1sin⁡(y)=sin⁡(y)+1sin⁡(y) 1 + \frac{1}{\sin(y)} = \frac{\sin(y)}{\sin(y)} + \frac{1}{\sin(y)} = \frac{\sin(y) + 1}{\sin(y)}
Now the expression becomes:
1+sin⁡(y)sin⁡(y)+1sin⁡(y) \frac{1+\sin(y)}{\frac{\sin(y) + 1}{\sin(y)}}

STEP 5

Simplify the overall expression by multiplying by the reciprocal of the denominator:
1+sin⁡(y)1+1sin⁡(y)=(1+sin⁡(y))×sin⁡(y)sin⁡(y)+1 \frac{1+\sin(y)}{1+\frac{1}{\sin(y)}} = (1+\sin(y)) \times \frac{\sin(y)}{\sin(y) + 1}
Notice that 1+sin⁡(y)1 + \sin(y) in the numerator and sin⁡(y)+1\sin(y) + 1 in the denominator are identical, so they cancel each other out:
1+sin⁡(y)sin⁡(y)+1=1 \frac{1+\sin(y)}{\sin(y) + 1} = 1
Thus, the simplified expression is:
sin⁡(y) \boxed{\sin(y)}

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