Question

Prove: sec(A + B)(cot A cot B - 1) = csc A csc B

Answer 1

To prove the identity sec(A + B)(cot A cot B - 1) = csc A csc B, we expand sec(A + B) using the sum formula for cosine and simplify (cot A cot B - 1) using trigonometric identities. Simplifying the left side shows us both sides of the equation are ultimately equal to 1, thus proving the identity.

Step-by-step explanation:

To prove that sec(A + B)(cot A cot B - 1) = csc A csc B, we will use trigonometric identities to simplify the left side of the equation to the right side.

We start by expanding sec(A + B) using the sum formula for cosine and then taking its reciprocal, since secant is the reciprocal of cosine:

1. sec(A + B) = 1 / cos(A + B)
2. cos(A + B) = cos A cos B - sin A sin B (Sum formula for cosine)
3. Therefore, sec(A + B) = 1 / (cos A cos B - sin A sin B)

Next, we'll simplify the term(cot A cot B - 1) by substituting cotangent with the reciprocal of tangent which is cos/sin:

1. cot A cot B - 1 = (cos A/sin A) * (cos B/sin B) - 1

Combining these gives us:

1. sec(A + B)(cot A cot B - 1) = [1 / (cos A cos B - sin A sin B)] * [(cos A cos B)/(sin A sin B) - 1]
2. Multiply the numerator of the second fraction by the denominator of the first fraction:
3. sec(A + B)(cot A cot B - 1) = [(cos A cos B) - (sin A sin B)] / (sin A sin B)

Notice that (cos A cos B) / (sin A sin B) simplifies to cot A cot B, and that sin^2 A + cos^2 A = 1 (Pythagorean identity). So, in terms of sine:

1. (sin^2 A + sin^2 B) = 1, therefore (cos A cos B) - (sin A sin B) = sin A sin B
2. Now the left side of the equation becomes sec(A + B)(cot A cot B - 1) = (sin A sin B) / (sin A sin B) = 1
3. Looking at the right side of the initial equation: csc A csc B = (1/sin A)(1/sin B) = 1

Both sides simplify to 1, so the identity sec(A + B)(cot A cot B - 1) = csc A csc B is proved.