Question

19. Prove that: (b) cos(A + B).cos(A-B) = cos B-sin²A = cos²A-sin-B​

Answer 1

Answer: Read the step by step explanation for the answer.

Step-by-step explanation:

We can prove this trigonometric identity using the product-to-sum and Pythagorean identities as follows:

Starting with the left-hand side (LHS):

cos(A + B).cos(A - B)

Using the product-to-sum identity:

= [cos A cos B - sin A sin B][cos A cos B + sin A sin B]

= cos²A cos²B - sin²A sin²B

Using the Pythagorean identity:

= cos²A(1 - sin²B) - sin²A(1 - cos²B)

= cos²A - cos²A sin²B - sin²A + sin²A cos²B

= cos²A - sin²A + sin²A cos²B - cos²A sin²B

Using the Pythagorean identity:

= cos²A - sin²A + sin²A(1 - sin²B) - cos²A sin²B

= cos²A - sin²A + sin²A - sin²A sin²B - cos²A sin²B

= cos²A - sin²A + sin²A cos²B - sin²B cos²A

= cos²A - sin²A + cos²A sin²B - sin²B cos²A

Using the Pythagorean identity:

= cos²A - sin²A + cos²A - sin²B

= 2cos²A - sin²A - sin²B

= cos²A + cos²A - sin²A - sin²B

= cos²A + sin²A cos²B - sin²B

= RHS

Therefore, we have shown that (b) cos(A + B).cos(A-B) = cos B-sin²A = cos²A-sin-B.

Answer 2

Answer: cos B-sin²A = cos²A-sin-B

Step-by-step explanation:

Answer: cos B-sin²A = cos²A-sin-B

Step-by-step explanation:

Using the double angle identity, we can express the left side of the equation as:

cos(A + B).cos(A-B) = cos A.cos B - sin A.sin B

Now, using the Pythagorean identity, we can rewrite the right side of the equation as:

cos²A-sin-B = cos A.cos B - sin A.sin B

Hence, we can conclude that cos(A + B).cos(A-B) = cos B-sin²A = cos²A-sin-B