A) \[\pm \frac{{{({{a}^{2}}+{{b}^{2}})}^{4}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\left( \frac{a}{{{b}^{8}}}+\frac{b}{{{a}^{8}}} \right)\]
B) \[\pm \frac{{{({{a}^{2}}+{{b}^{2}})}^{4}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\left( \frac{a}{{{b}^{8}}}-\frac{b}{{{a}^{8}}} \right)\]
C) \[\pm \frac{{{({{a}^{2}}-{{b}^{2}})}^{4}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\left( \frac{a}{{{b}^{8}}}+\frac{b}{{{a}^{8}}} \right)\]
D) \[\pm \frac{{{({{a}^{2}}-{{b}^{2}})}^{4}}}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\left( \frac{a}{{{b}^{8}}}-\frac{b}{{{a}^{8}}} \right)\]
Correct Answer: A
Solution :
Given that \[\tan \theta =\frac{a}{b}\] and \[\cos \,2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\frac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}}\] \[\sin \theta =\pm \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}},\,\,\cos \,\theta =\pm \frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] Now, \[\frac{\sin \,\theta }{\cos {{\,}^{8}}\theta }+\frac{\cos \,\theta }{{{\sin }^{8}}\theta }=\frac{\left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}{{{\left( \frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}^{8}}}+\frac{\left( \frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}{{{\left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}^{8}}}\] \[=\frac{a\,{{({{a}^{2}}+{{b}^{2}})}^{4}}}{{{b}^{8}}\,{{({{a}^{2}}+{{b}^{2}})}^{1/2}}}+\frac{b\,{{({{a}^{2}}+{{b}^{2}})}^{4}}}{{{a}^{8}}\,{{({{a}^{2}}+{{b}^{2}})}^{1/2}}}\] \[=\pm \frac{{{({{a}^{2}}+{{b}^{2}})}^{4}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\,\left( \frac{a}{{{b}^{8}}}+\frac{b}{{{a}^{8}}} \right)\].
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