Question
In a triangle $$ABC,DC = {90^ \circ }$$ then $$\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}$$ is equal to :
A.
$$\sin \left( {A + B} \right)$$
B.
$$\sin \left( {A - B} \right)$$
C.
$$\cos \left( {A + B} \right)$$
D.
$$\sin \left( {\frac{{A - B}}{2}} \right)$$
Answer :
$$\sin \left( {A - B} \right)$$
Solution :
$$\eqalign{
& A + B = {180^ \circ } - C = {90^ \circ } \cr
& a = 2R\sin A,b = 2R\sin B,c = 2R\sin C \cr
& \therefore \frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}} = \frac{{{{\sin }^2}A - {{\sin }^2}B}}{{{{\sin }^2}A + {{\sin }^2}B}} \cr
& = \frac{{\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{{{\sin }^2}A + {{\sin }^2}\left( {{{90}^ \circ } - A} \right)}}\,\,\,\left[ {\because A + B = {{90}^ \circ }} \right] \cr
& = \frac{{\sin {{90}^ \circ }\sin \left( {A - B} \right)}}{{{{\sin }^2}A + {{\cos }^2}A}} = \sin \left( {A - B} \right) \cr} $$