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

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Question

(A) Given that $	an 	heta = \frac{a}{b}$.Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{a}{b}$,we can write $\sin 	heta = \pm \frac{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 that $	an 	heta = \frac{a}{b}$.Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{a}{b}$,we can write $\sin 	heta = \pm \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos 	heta = \pm \frac{b}{\sqrt{a^2 + b^2}}$.Now,consider the expression $\frac{\sin 	heta}{\cos^8 	heta} + \frac{\cos 	heta}{\sin^8 	heta}$.Substituting the values of $\sin 	heta$ and $\cos 	heta$:$= \frac{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} ight)}{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} ight)^8} + \frac{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} ight)}{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} ight)^8}$$= \frac{\pm a (a^2 + b^2)^4}{b^8 \sqrt{a^2 + b^2}} + \frac{\pm b (a^2 + b^2)^4}{a^8 \sqrt{a^2 + b^2}}$$= \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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