If tan theta = frac{a}{b}, then frac{sin thet | vedclass

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(A) Given $	an 	heta = \frac{a}{b}.$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{a}{b},$ we have $\sin 	heta = \pm \frac{a}{\sqrt{a

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $	an 	heta = \frac{a}{b}.$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{a}{b},$ we have $\sin 	heta = \pm \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos 	heta = \pm \frac{b}{\sqrt{a^2 + b^2}}.$Now,consider the expression $E = \frac{\sin 	heta}{\cos^8 	heta} + \frac{\cos 	heta}{\sin^8 	heta}.$Substituting the values of $\sin 	heta$ and $\cos 	heta$:$E = \frac{\pm \frac{a}{\sqrt{a^2 + b^2}}}{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} ight)^8} + \frac{\pm \frac{b}{\sqrt{a^2 + b^2}}}{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} ight)^8}$Since $(\pm x)^8 = x^8,$ the signs simplify to $\pm$ for the whole expression:$E = \pm \left[ \frac{a \cdot (a^2 + b^2)^4}{b^8 \cdot \sqrt{a^2 + b^2}} + \frac{b \cdot (a^2 + b^2)^4}{a^8 \cdot \sqrt{a^2 + b^2}} ight]$$E = \pm \frac{(a^2 + b^2)^4}{\sqrt{a^2 + b^2}} \left( \frac{a}{b^8} + \frac{b}{a^8} ight).$

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

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