A) \[\sin (A+B)\]
B) \[\sin (A-B)\]
C) \[\cos (A+B)\]
D) \[\sin \left( \frac{A-B}{2} \right)\]
Correct Answer: A
Solution :
[b] \[A+B=180{}^\circ -C=90{}^\circ \] \[a=2R\sin A,\,\,b=2R\sin B,\,\,c=2R\sin C\] \[\therefore \,\,\,\,\,\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}B}\] \[=\frac{\sin (A+B)sin(A-B)}{{{\sin }^{2}}A+{{\sin }^{2}}(90{}^\circ -A)}\] \[[\because \,\,\,A+B=90{}^\circ ]\] \[=\frac{\sin 90{}^\circ \sin (A-B)}{{{\sin }^{2}}A+{{\cos }^{2}}A}=\sin (A-B)\]
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