Trigonometric Identities Question 173
Question: If $ \tan \theta =\frac{a}{b}, $ then $ \frac{\sin \theta }{{{\cos }^{8}}\theta }+\frac{\cos \theta }{{{\sin }^{8}}\theta }= $
[WB JEE 1986]
Options:
A) $ \pm \frac{{{(a^{2}+b^{2})}^{4}}}{\sqrt{a^{2}+b^{2}}}( \frac{a}{b^{8}}+\frac{b}{a^{8}} ) $
B) $ \pm \frac{{{(a^{2}+b^{2})}^{4}}}{\sqrt{a^{2}+b^{2}}}( \frac{a}{b^{8}}-\frac{b}{a^{8}} ) $
C) $ \pm \frac{{{(a^{2}-b^{2})}^{4}}}{\sqrt{a^{2}+b^{2}}}( \frac{a}{b^{8}}+\frac{b}{a^{8}} ) $
D) $ \pm \frac{{{(a^{2}-b^{2})}^{4}}}{\sqrt{a^{2}-b^{2}}}( \frac{a}{b^{8}}-\frac{b}{a^{8}} ) $
Show Answer
Answer:
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{( \frac{a}{\sqrt{a^{2}+b^{2}}} )}{{{( \frac{b}{\sqrt{a^{2}+b^{2}}} )}^{8}}}+\frac{( \frac{b}{\sqrt{a^{2}+b^{2}}} )}{{{( \frac{a}{\sqrt{a^{2}+b^{2}}} )}^{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}}},( \frac{a}{b^{8}}+\frac{b}{a^{8}} ) $ .
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