A) \[\frac{{{a}^{4}}+{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]
B) \[\frac{{{a}^{4}}-{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]
C) \[\frac{{{a}^{4}}-{{b}^{4}}+4{{a}^{2}}}{{{a}^{2}}{{b}^{2}}+{{a}^{4}}}\]
D) None of the above
Correct Answer: B
Solution :
\[\sin x+\sin y=a\] \[\Rightarrow \,\,2\sin \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)=a\] ...(1) \[\cos x+\cos y=b\] \[\Rightarrow \,\,\,\,2\cos \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)=b\] ...(2) dividing eq. (1) & (2) \[\tan \left( \frac{x+y}{2} \right)=\frac{a}{b}\] Squaring of eq. (1) & (2) and adding - \[4{{\cos }^{2}}\left( \frac{x-y}{2} \right)={{a}^{2}}+{{b}^{2}}\] \[se{{c}^{2}}\left( \frac{x-y}{2} \right)=\frac{4}{{{a}^{2}}+{{b}^{2}}}\] again- \[{{\tan }^{2}}\left( \frac{x+y}{2} \right)+{{\tan }^{2}}\left( \frac{x-y}{2} \right)\] \[={{\left( \frac{a}{b} \right)}^{2}}+{{\sec }^{2}}\left( \frac{x-y}{2} \right)-1\] \[=\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{4}{{{a}^{2}}+{{b}^{2}}}-1\] \[=\frac{{{a}^{4}}-{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]
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