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

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Question

(a) Given that $	an 	heta = \frac{a}{b}$ 

and $\cos \,2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }} = \frac{{{b^2} - {

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}$ 

and $\cos \,2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }} = \frac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}$

$	herefore $ $\sin 	heta = \pm \frac{a}{{\sqrt {{a^2} + {b^2}} }},\,\,\cos \,	heta = \pm \frac{b}{{\sqrt {{a^2} + {b^2}} }}$

Now, $\frac{{\sin \,	heta }}{{\cos {\,^8}	heta }} + \frac{{\cos \,	heta }}{{{{\sin }^8}	heta }}$

$= \frac{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} ight)}}{{{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} ight)}^8}}} + \frac{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} ight)}}{{{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} ight)}^8}}}$

$ = \frac{{a\,{{({a^2} + {b^2})}^4}}}{{{b^8}\,{{({a^2} + {b^2})}^{1/2}}}} + \frac{{b\,{{({a^2} + {b^2})}^4}}}{{{a^8}\,{{({a^2} + {b^2})}^{1/2}}}}$

$ = \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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