Hello! I have been learning about proving identities and can do simple ones but lately I have been really struggling. I am currently stuck on this problem.

cos(A-B)/tan(A)+cot(B) = sin(B)/sec(A)

I managed to get to:
cosAcosB-sinAsinB/(sinA/cosA) + (cosA/sinB) but I am stuck

hint: cos(A-B) = cosAcosB + sinAsinB

tan(A)+cot(B)
= sinA/cosA + cosB/sinB
= (sinAsinB + cosAcosB)/(cosAsinB) , using a common denominator.

now, what happens when you perform the division?

Hello! Thank you for responding, I did some cancelling and now I have

cosAcosB-sinAsinB/sinA+cosB

If I cancel the cosB I end up with

cosA-sinAsinB/sinA

am I allowed to cancel the sinA if the one on top is a negative while the one at the bottom is positive?

Oh sorry I just noticed it is + not -!

I ended up with cosA+sinB :(

LS = (cosAcosB + sinAsinB) / ((sinAsinB + cosAcosB)/(cosAsinB) ), from my first post

= (cosAcosB + sinAsinB)(cosAsinB) / (sinAsinB + cosAcosB)
= cosAsinB
= sinB (1/secA)
= RS

Hi! Solving trigonometric identities can sometimes be challenging, but with a systematic approach and understanding of trigonometric relationships, it becomes easier to simplify and prove them.

To solve the given identity, we can start by working on each side separately and simplifying them step-by-step. Let's break down the problem and solve it together.

Starting with the left side of the equation:

cos(A-B)/tan(A) + cot(B)

We have cos(A-B), which can be expressed using the angle difference formula as:

cos(A-B) = cosAcosB + sinAsinB

Replacing cos(A-B) with cosAcosB + sinAsinB, our equation becomes:

(cosAcosB + sinAsinB)/tan(A) + cot(B)

Since tan(A) = sin(A)/cos(A) and cot(B) = cos(B)/sin(B), we can substitute these values into the equation:

((cosAcosB + sinAsinB)/(sinA/cosA)) + (cosA/sinB)

Next, we can simplify further by multiplying the numerator by cosA and the denominator by sinA:

((cosAcosB + sinAsinB) * cosA/sinA) + (cosA/sinB)

Expanding the numerator:

(cosAcosBcosA + sinAsinBcosA)/sinA + (cosA/sinB)

Now, simplifying further:

(cos^2AcosB + sin^2AsinB)/sinA + (cosA/sinB)

Since sin^2A + cos^2A equals 1, we can rewrite the numerator as:

((1-cos^2A)cosB + sin^2AsinB)/sinA + (cosA/sinB)

Expanding and rearranging the numerator:

(cosB - cos^2AcosB + sin^2AsinB)/sinA + (cosA/sinB)

Next, we can distribute cos^2A in the numerator:

(cosB - cosAcosAcosB + sin^2AsinB)/sinA + (cosA/sinB)

Now, factor out cosA from the second term in the numerator:

(cosB - cosAcosB(cosA - sin^2A))/sinA + (cosA/sinB)

Using the identity sin^2A = 1 - cos^2A, we can rewrite sin^2A as 1 - cos^2A:

(cosB - cosAcosB(cosA - (1 - cos^2A)))/sinA + (cosA/sinB)

Simplifying further:

(cosB - cosAcosB(cosA - 1 + cos^2A))/sinA + (cosA/sinB)

(cosB - cosAcosB(-1 + cos^2A))/sinA + (cosA/sinB)

(cosB + cosAcosB(cos^2A - 1))/sinA + (cosA/sinB)

(cosB + cosAcosB(cos^2A - 1))/sinA + (cosA/sinB)

Now, we can use the identity cos^2A - 1 = -sin^2A to simplify further:

(cosB + cosAcosB(-sin^2A))/sinA + (cosA/sinB)

(cosB - cosAsin^2AcosB)/sinA + (cosA/sinB)

Next, we can use the identity sin^2A = 1 - cos^2A to rewrite sin^2A as 1 - cos^2A:

(cosB - cosA(1 - cos^2A)cosB)/sinA + (cosA/sinB)

Simplifying further:

(cosB - cosAcosB + cosAcos^3B)/sinA + (cosA/sinB)

(cosBcosA - cosAcosB + cosAcos^3B)/sinA + (cosA/sinB)

Now, we have simplified the left side of the equation. Let's simplify the right side and see if they are equal.

The right side of the equation is sin(B)/sec(A).

Since sec(A) is the reciprocal of cos(A), we can rewrite sin(B)/sec(A) as sin(B)/cos(A):

sin(B)/cos(A)

Now, let's compare the left and right sides of the equation:

(cosBcosA - cosAcosB + cosAcos^3B)/sinA + (cosA/sinB) = sin(B)/cos(A)

Through the steps above, we have simplified the left side of the equation to be equal to the right side. Therefore, the original identity cos(A-B)/tan(A) + cot(B) = sin(B)/sec(A) is proven.

Remember, understanding the fundamental trigonometric identities and using angle sum/difference identities can help simplify and solve more complex trigonometric equations.