Answer:
Step-by-step explanation:
We can simplify the expression (1-cos α)(1+sec α) using trigonometric identities. Here's how:
Rewrite sec α:
We know that sec α is 1/cos α. Substitute this into the expression:
(1-cos α)(1 + (1/cos α))
Expand the bracket:
(1-cos α) * (1 + 1/cos α) = 1 + 1/cos α - cos α - cos^2(α)
Simplify:
Combine terms with cos α:
1 + 1/cos α - cos α - cos^2(α) = 1 - cos^2(α) + 1/cos α
We recognize the first part (1 - cos^2(α)) as another trigonometric identity: sin^2(α).
Apply the identity (optional but clarifies the answer):
1 - cos^2(α) = sin^2(α) --> Substitute this identity
sin^2(α) + 1/cos α
Note: Since the question states the answer is not tan α or sec α - cos α, we cannot simply stop here.
Analyze 1/cos α:
We can try to manipulate 1/cos α further using other trigonometric identities. One possibility is to rewrite it using the Pythagorean identity (sin^2(α) + cos^2(α) = 1) and isolate cos α. However, this wouldn't lead to a simpler form that doesn't include sec α or tan α.
Therefore, the most simplified form we can achieve without introducing other trigonometric functions is:
sin^2(α) + 1/cos α
Answer:
[tex]\tan\alpha\sin\alpha[/tex]
Step-by-step explanation:
Given trigonometric expression:
[tex](1-\cos \alpha)(1+\sec \alpha)[/tex]
Since the secant function is the reciprocal of the cosine function, we can express sec(α) in terms of cos(α):
[tex](1-\cos \alpha)\left(1+\dfrac{1}{\cos \alpha}\right)[/tex]
Now, expand the brackets:
[tex]1+\dfrac{1}{\cos \alpha}-\cos \alpha -\dfrac{\cos \alpha}{\cos \alpha}[/tex]
Simplify:
[tex]1+\dfrac{1}{\cos \alpha}-\cos \alpha -1\\\\\\\\\dfrac{1}{\cos \alpha}-\cos \alpha[/tex]
Rewrite cos(α) as cos²(α) / cos(α) :
[tex]\dfrac{1}{\cos \alpha}-\dfrac{\cos^2 \alpha}{\cos \alpha}[/tex]
Now that the two fractions have the same denominator, combine the numerators:
[tex]\\\\\\\dfrac{1-\cos^2 \alpha}{\cos \alpha}[/tex]
Using the trigonometric identity sin²θ + cos²θ = 1, rewrite 1 - cos²(α) as sin²(α):
[tex]\dfrac{\sin^2\alpha}{\cos \alpha}[/tex]
Rewrite:
[tex]\sin\alpha \cdot \dfrac{\sin\alpha}{\cos \alpha}[/tex]
Since sin(α) / cos(α) = tanα, this simplifies to:
[tex]\LARGE\boxed{\boxed{\tan\alpha\sin\alpha}}[/tex]
