Trigonometric identity with sin, cos, and tan Very cool problem I was reading today!
$(1)$ $2\sin^2 A $ – $ 2\tan A \sin A + 2 $ – $ \sin^2 A $ – $ \cos^2 B = 0; $
$(2)$ $2\cos^2 B $ – $ 2\tan A \cos B + 2 $ – $ \sin^2 A $ - $ \cos^2 B = 0;$
If $\sin A ≠ \cos B$, find the value of $\boxed{\sin^2 A}$
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I realized there was an easy simplification in equation $(1)$; $2 \sin^2{A} - \sin^2{A} = \sin^2{A}$
There was another easy simplification in equation $(2); 2\cos^2{B} - \cos^2{B} = \cos^2{B}$
After simplifying $(1)$ and $(2)$, I got
$(3)$ $\sin^2 A $ – $ 2\tan A \sin A + 2 $ $ – \cos^2 B = 0$;
And $(4)$ $\cos^2 B $ – $ 2\tan A \cos B + 2 $ – $ \sin^2 A = 0$
Adding $(3)$ and $(4)$ gives
$(5)$ $- 2\tan A \sin A + 2\tan A \cos B + 4 = 0$ which can also be written as $\tan{A}(\sin{A} - \cos{B}) = 2$
Subtracting $(3)$ and $(4)$ gives
$(6)$ $2\sin^2{A} - 2\cos^2{B} - 2\tan{A}(\sin A - \cos{B}) = 0$
Substituting $(5)$ into $(6)$ gives
$(7)$ $2\sin^2{A} - 2\cos^2{B} - 4 = 0$
I am not really sure what to do after this. Could someone please explain what I should do? The $\sin{A}$ and $\cos{B}$ are very different and I can't seem to use any identities to solve the equation.

You have a sign error in (5), it should be $-2\tan A \cos B$. Thus, $$ \tan A(\sin A+ \cos B)=2$$ and $$\sin^2 A-\cos^2 B=\tan A (\sin A - \cos B) \implies \sin A+ \cos B=\tan A$$ Can you finish now?
