Question

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B​

Answer 1

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.