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

Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

trigonometry