<s>[INST] In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then $C=?$ In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then  $C=?$
We have $\sin^2{A}+\cos^2{C}=\cos^2{B}  
\implies 2\sin^2{A}+2\cos^2{C}=2\cos^2{B}  
\implies 1-\cos{2A}+\cos{2C}-1=\cos{2B}-1  
\implies 1+\cos{2C}=\cos{2B}+\cos{2A}  
 \implies 1+\cos{2C}=2\cos(A+B)\cos(A-B)   
 \implies 1+\cos{2C}=2\cos(\pi-A)\cos(A-B)$, since $A+B+C=\pi$
 $\implies 1+\cos{2C}=2\cos{A}\cos(A-B)$  
How can I find $C$ ? Please help.
 [/INST] Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin^2A=\cos^2B-\cos^2C=\sin^2C-\sin^2B=\sin(C+B)\sin(C-B)$$
$$\implies\sin A\sin(\pi-\overline{B+C})=\sin(\pi-A)\sin(C-B)$$
$$\implies\sin A[\sin(B+C)-\sin(C-B)]=0$$
$$\implies\sin A[2\sin B\cos C]=0$$
$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)

Alternatively, $$\cos^2C=\cos^2B-\sin^2A=\cos(B+A)\cos(B-A)$$
Again, $\cos(A+B)=\cos(\pi-C)=-\cos C$
$$\implies\cos C[-\cos(A+B)]=-\cos C\cos(B-A)\iff\cos C[\cos(B-A)-\cos(B+A)]=0$$
$$\iff\cos C[2\sin B\sin A]=0$$
$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)
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