A) \[m{{(m{{n}^{2}})}^{1/3}}-n{{(n{{m}^{2}})}^{1/3}}=1\]
B) \[m{{({{m}^{2}}n)}^{1/3}}-n{{(m{{n}^{2}})}^{1/3}}=1\]
C) \[n{{(m{{n}^{2}})}^{1/3}}-m{{(n{{m}^{2}})}^{1/3}}=1\]
D) \[n{{({{m}^{2}}n)}^{1/3}}-m{{(m{{n}^{2}})}^{1/3}}=1\]
Correct Answer: A
Solution :
As given \[\frac{1}{\tan \theta }+\tan \theta =m\,\Rightarrow \,1+{{\tan }^{2}}\theta =m\,\tan \theta \] \[\Rightarrow \,\,{{\sec }^{2}}\theta =m\,\tan \theta \] ?..(i) and \[\sec \theta -\cos \theta =n\,\,\Rightarrow \,\,{{\sec }^{2}}\theta -1=n\,\sec \theta \] \[\Rightarrow \,\,{{\tan }^{2}}\theta =n\,\,\sec \theta \] \[\Rightarrow \,\,{{\tan }^{4}}\theta ={{n}^{2}}\,{{\sec }^{2}}\theta ={{n}^{2}}.\,m\,\,\tan \theta \] {by (i)} \[\Rightarrow \,\,{{\tan }^{3}}\theta ={{n}^{2}}m\,,\,\,\,(\,\because \,\,\tan \theta \ne 0)\] \[\Rightarrow \,\,\tan \theta ={{({{n}^{2}}m)}^{1/3}}\] ?..(ii) Also, \[{{\sec }^{2}}\theta =m\,\,\tan \theta =m\,{{({{n}^{2}}m)}^{1/3}}\] {by (i) and (ii)} \ Using the identity \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] \[\Rightarrow \,\,m\,{{(m{{n}^{2}})}^{1/3}}-{{({{n}^{2}}m)}^{2/3}}=1\] \[\Rightarrow \,\,m\,{{(m{{n}^{2}})}^{1/3}}-n\,{{(n{{m}^{2}})}^{1/3}}=1.\]
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