A) \[x=\frac{y+1}{y-1}\]
B) \[x=\frac{y-1}{y+1}\]
C) \[y=\frac{1-x}{1+x}\]
D) None of these
Correct Answer: B
Solution :
We have \[xy=(\sec \varphi -\tan \varphi )\,\,\text{(cosec}\,\,\varphi +\cot \,\,\varphi )\] \[=\frac{1-\sin \,\varphi }{\cos \,\varphi }\,.\,\frac{1+\cos \,\varphi }{\sin \,\varphi }\] \[\Rightarrow \,xy+1=\frac{1-\sin \,\varphi +\cos \,\varphi -\sin \,\varphi \,\cos \,\varphi +\sin \varphi \cos \varphi }{\cos \varphi \sin \varphi }\] \[=\frac{1-\sin \,\varphi +\cos \,\varphi }{\cos \,\varphi \sin \,\varphi }\] ?..(i) \[x-y=(\sec \,\varphi -\tan \,\varphi )-(\cos ec\,\varphi +\cot \,\varphi )\] \[=\frac{1-\sin \,\varphi }{\cos \,\varphi }-\frac{1+\cos \,\varphi }{\sin \,\varphi }=\frac{\sin \,\varphi -{{\sin }^{2}}\varphi -\cos \,\varphi -{{\cos }^{2}}\varphi }{\cos \,\varphi \,\sin \,\varphi }\] \[=\frac{\sin \,\varphi -\cos \,\varphi -1}{\cos \,\varphi \,\sin \,\varphi }\] ?..(ii) Adding (i) and (ii) we get, \[xy+1+(x-y)=0\] \[\Rightarrow x=\frac{y-1}{y+1}\].
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