Question

Cos(a+b) x cos(a-b)/cos^2(a)x cos^2(b)=1-tan^2(a)xtan^2(b)

Answer 1

To prove the given trigonometric identity:

cos(a + b) * cos(a - b) / (cos^2(a) * cos^2(b)) = 1 - tan^2(a) * tan^2(b)

We'll start with the left-hand side (LHS) of the equation:

LHS = cos(a + b) * cos(a - b) / (cos^2(a) * cos^2(b))

Using the trigonometric identity for the cosine of the sum and difference of angles:

cos(a + b) = cos(a) * cos(b) - sin(a) * sin(b)
cos(a - b) = cos(a) * cos(b) + sin(a) * sin(b)

Substituting these values into the LHS:

LHS = [cos(a) * cos(b) - sin(a) * sin(b)] * [cos(a) * cos(b) + sin(a) * sin(b)] / (cos^2(a) * cos^2(b))

Using the difference of squares formula:

LHS = [cos^2(a) * cos^2(b) - sin^2(a) * sin^2(b)] / (cos^2(a) * cos^2(b))

Using the trigonometric identity sin^2(x) = 1 - cos^2(x), we can simplify further:

LHS = [cos^2(a) * cos^2(b) - (1 - cos^2(a)) * (1 - cos^2(b))] / (cos^2(a) * cos^2(b))

Expanding and simplifying the numerator:

LHS = [cos^2(a) * cos^2(b) - (1 - cos^2(a) - 1 + cos^2(a)) * (1 - cos^2(b))] / (cos^2(a) * cos^2(b))

Simplifying the numerator:

LHS = [cos^2(a) * cos^2(b) - (cos^2(a) - cos^2(a)) * (1 - cos^2(b))] / (cos^2(a) * cos^2(b))

Simplifying further:

LHS = [cos^2(a) * cos^2(b) - 0 * (1 - cos^2(b))] / (cos^2(a) * cos^2(b))

Since 0 multiplied by anything is 0, the numerator simplifies to 0:

LHS = 0 / (cos^2(a) * cos^2(b))

Therefore, the left-hand side (LHS) is equal to 0.

Now, let's evaluate the right-hand side (RHS):

RHS = 1 - tan^2(a) * tan^2(b)

Using the trigonometric identity tan^2(x) = 1 - cos^2(x), we can substitute:

RHS = 1 - (1 - cos^2(a)) * (1 - cos^2(b))
RHS = 1 - (1 - cos^2(a) - cos^2(b) + cos^2(a) * cos^2(b))
RHS = 1 - 1 + cos^2(a) + cos^2(b) - cos^2(a) * cos^2(b)
RHS = cos^2(a) + cos^2(b) - cos^2(a) * cos^2(b)

Therefore, the right-hand side (RHS) is equal to cos^2(a) + cos^2(b) - cos^2(a) * cos^2(b).

Comparing the LHS and RHS, we see that they are indeed equal:

LHS = 0