Question

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

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

Answer 1

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

We have $\tan \theta =\frac{a}{b}=\frac{oppositeside}{adjacentside}$

$\therefore$by Pythagoras theorem, we have ${\left(hypotenuse\right)}^2={\left(oppositeside\right)}^2+{\left(adjacentside\right)}^2=\sqrt{a^2+b^2}$

$\tan \theta$ is positive in first and third quadrants.

When $\theta$ is in first quadrant,we have

$\sin \theta =\frac{a}{\sqrt{a^2+b^2}}$

and $\cos \theta =\frac{b}{\sqrt{a^2+b^2}}$

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

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

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

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

When $\theta$ is in third quadrant,we have

$\sin \theta =\frac{-a}{\sqrt{a^2+b^2}}$

and $\cos \theta =\frac{-b}{\sqrt{a^2+b^2}}$

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

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

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

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

$\therefore \frac{\sin \theta }{{\cos }^8\theta }+\frac{\cos \theta }{{\sin }^8\theta }=\pm {(a^2+b^2)}^{4-\frac{1}{2}}(\frac{a}{b^8}+\frac{b}{a^8})$

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