Question

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

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

Answer 1

The correct option is A $\pm \frac{{(a^2+b^2)}^4}{\sqrt{a^2+b^2}}(\frac{a}{b^8}+\frac{b}{a^8})$

Given that,

$\tan \theta =\frac{a}{b}$

$\frac{\sin \theta }{{\cos }^8\theta }+\frac{\cos \theta }{{\sin }^8\theta }$

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

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

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

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

Part $A$ is correct answer.