Question

If $\tan\theta = \frac{a}{b},$ then $\frac{\sin\theta}{\cos^{8}\theta} + \frac{\cos\theta}{\sin^{8}\theta} =$

• 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. $\pm \frac{(a^{2} - b^{2})^{4}}{\sqrt{a^{2} - b^{2}}}\left( \frac{a}{b^{8}} - \frac{b}{a^{8}} \right)$

Answer 1

Answer: (A)

$\pm \frac{(a^{2} + b^{2})^{4}}{\sqrt{a^{2} + b^{2}}}\left( \frac{a}{b^{8}} + \frac{b}{a^{8}} \right)$

Answer 2

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, $\frac{\sin\theta}{\cos^{8}\theta} + \frac{\cos\theta}{\sin^{8}\theta} = \frac{\left( \frac{a}{\sqrt{a^{2} + b^{2}}} \right)}{\left( \frac{b}{\sqrt{a^{2} + b^{2}}} \right)^{8}} + \frac{\left( \frac{b}{\sqrt{a^{2} + b^{2}}} \right)}{\left( \frac{a}{\sqrt{a^{2} + b^{2}}} \right)^{8}}$

$= \frac{a(a^{2} + b^{2})^{4}}{b^{8}(a^{2} + b^{2})^{1/2}} + \frac{b(a^{2} + b^{2})^{4}}{a^{8}(a^{2} + b^{2})^{1/2}}$

$= \pm \frac{(a^{2} + b^{2})^{4}}{\sqrt{a^{2} + b^{2}}}\left( \frac{a}{b^{8}} + \frac{b}{a^{8}} \right)$.