Question

(cosA/1-sinA) - (cosA/1+sinA)

Answer 1

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Answer 2

To simplify the expression (cosA/1-sinA) - (cosA/1+sinA), we can first find a common denominator for both terms.

For the first term, (cosA/1-sinA), we can multiply the numerator and denominator by (1+sinA):

(cosA/1-sinA) * (1+sinA)/(1+sinA)

Expanding this, we get:

(cosA * (1+sinA)) / ((1-sinA) * (1+sinA))

Simplifying further, we have:

(cosA + cosA * sinA) / (1 - sin^2 A)

Recall the identity sin^2 A + cos^2 A = 1. Therefore, we can substitute 1 - sin^2 A with cos^2 A:

(cosA + cosA * sinA) / cos^2 A

Now, let's simplify the second term, (cosA/1+sinA). Similarly, we can multiply the numerator and denominator by (1-sinA):

(cosA/1+sinA) * (1-sinA)/(1-sinA)

Expanding this, we get:

(cosA * (1-sinA)) / ((1+sinA) * (1-sinA))

Simplifying further, we have:

(cosA - cosA * sinA) / (1 - sin^2 A)

Again, we can substitute 1 - sin^2 A with cos^2 A:

(cosA - cosA * sinA) / cos^2 A

Now, subtracting the two terms, we have:

(cosA + cosA * sinA) / cos^2 A - (cosA - cosA * sinA) / cos^2 A

Since the denominators are the same, we can combine the numerators:

(cosA + cosA * sinA - cosA + cosA * sinA) / cos^2 A

Simplifying this further, we get:

(2cosA * sinA) / cos^2 A

The cosA terms cancel out, leaving us with:

2sinA / cos A

Finally, we can simplify further by dividing both the numerator and denominator by cos A:

2tan A

Therefore, (cosA/1-sinA) - (cosA/1+sinA) simplifies to 2tan A.