Prove this identity

[jwpaine]

Prove that:
\(\displaystyle \L tan(x-y) = \frac{cot(y)\,-cot(x)}{(cot(x) \cdot cot(y)) +1}\)




This is what I have so far.... I'll start with the right side:

\(\displaystyle \L \frac{\frac{cos(x)}{sin(y)}\,-\,\frac{cos(x)}{sin(x)}}{\frac{cos(x)cos(y)\,+\,sin(x)sin(y)}{sin(x)sin(y) } }\)

re-write the division as multiplication by the reciprocal.

\(\displaystyle \L \frac{cox(y)sin(x)\,-\,cos(x)sin(y)}{sin(y)(sin(x)}\,\cdot\,\frac{sin(x)sin(y)}{cos(x)cos(y)\,+\,sin(x)sin(y))}\)

Multiply both sides and factor:

\(\displaystyle \L \frac{sin(y)(sin(y)cos(y)\,-\,sin(x)cos(x))sin(y)}{sin(x)(cos(x)(cos(y)\,+\,sin(x)sin(y))sin(y)}\)

Where do I go from here? I might have messed up somewhere... and oh my god is this getting complicated.

Thanks,
John.

 

[Random]

You are forgetting an identity

See where to go from there? Start over, it will probably be easier.

 

[jwpaine]

Thank you very much!

 

[soroban]

Hello, John!

You were doing fine . . .

Prove that: \(\displaystyle \L\:\tan(x\,-\,y) \:= \:\frac{\cot(y)\,-\,\cot(x)}{\cot(x)\cdot\cot(y))\,+\,1}\)

This is what I have so far.... I'll start with the right side:

\(\displaystyle \L \frac{\frac{\cos(x)}{\sin(y)}\,-\,\frac{\cos(x)}{\sin(x)}}{\frac{\cos(x)\cos(y)\,+\,\sin(x)\sin(y)}{\sin(x)\sin(y)}}\)


Re-write the division as multiplication by the reciprocal:

\(\displaystyle \L\frac{\cos(y)\sin(x)\,-\,\cos(x)\sin(y)}{\sin(y)\sin(x)}\,\cdot\,\frac{\sin(x)\sin(y)}{\cos(x)\cos(y)\,+\,\sin(x)\sin(y)}\;\) . . . Good!

Multiply: \(\displaystyle \L\:\frac{\sin(x)\sin(y)\cdot\overbrace{[sin(x)cos(y)\,-\,sin(y)cos(x)]}^{\text{This is }\sin(x-y)}}{sin(x)\sin(y)\cdot\underbrace{[cos(x)cos(y)\,+\,sin(x)sin(y)]}_{\text{This is {cos(x-y)}}\)

Therefore: \(\displaystyle \L\:\frac{\sin(x\,-\,y)}{\cos(x\,-\,y)}\)\(\displaystyle \;=\;\tan(x\,-\,y)\)