Sec(a-B) = (cota cotB+1)/(1+tanatanB)

[CrossFit415]

Homework Statement

sec(a-b) = (cot a cot B+1)/(1+tan a tan B)

so I got 1/cos a (1/cos B) -sin a (sin b) but they got cot a cot B + 1 / 1+tan a tan B

What formula should I use since there's no sec formula?

Homework Equations

The Attempt at a Solution

 

[vela]

If you're trying to verify that the expression is an identity, don't bother. It's not. For a=pi/3 and b=pi/6, for instance, you get sec(a-b)=2/sqrt(3) while the RHS is equal to 1.

 

[CrossFit415]

If you're trying to verify that the expression is an identity, don't bother. It's not. For a=pi/3 and b=pi/6, for instance, you get sec(a-b)=2/sqrt(3) while the RHS is equal to 1.

I don't understand? Then how did they get that? Thanks

 

[vela]

They made a mistake somewhere. The problem itself is wrong.

 

[CrossFit415]

whoops lol. I made a mistake! sorry! the answer is ( sec a sec B ) / 1 + tan a tan B

 

[D H]

Now that is an identity, or at least it will be after you get your parentheses correct.

 

[CrossFit415]

Hmm.. So Sec (a - B) would expand to 1/cos (a - B) ? or would I have to use tan (a - B)? Thank you.

 

[eumyang]

I would start with the right-hand side instead, and multiply numerator and denominator by cos A cos B:
[tex]\frac{\sec A \sec B}{1 + \tan A \tan B} \cdot \frac{\cos A \cos B}{\cos A \cos B} = ...[/tex]
Can you take it from there?

 

[CrossFit415]

I would start with the right-hand side instead, and multiply numerator and denominator by cos A cos B:
[tex]\frac{\sec A \sec B}{1 + \tan A \tan B} \cdot \frac{\cos A \cos B}{\cos A \cos B} = ...[/tex]
Can you take it from there?

Hmm, never thought about that, thanks.