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यदि $\tan \theta = \frac{a}{b},$ तो $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $
$ \pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$
$ \pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} - \frac{b}{{{a^8}}}} \right)$
$ \pm \frac{{{{({a^2} - {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$
$ \pm \frac{{{{({a^2} - {b^2})}^4}}}{{\sqrt {{a^2} - {b^2}} }}\left( {\frac{a}{{{b^8}}} - \frac{b}{{{a^8}}}} \right)$
Solution
(a) दिया है $\tan \theta = \frac{a}{b}$
एवं $\cos \,2\theta = \frac{{1 – {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \frac{{{b^2} – {a^2}}}{{{b^2} + {a^2}}}$
$\therefore $ $\sin \theta = \pm \frac{a}{{\sqrt {{a^2} + {b^2}} }},\,\,\cos \,\theta = \pm \frac{b}{{\sqrt {{a^2} + {b^2}} }}$
अब, $\frac{{\sin \,\theta }}{{\cos {\,^8}\theta }} + \frac{{\cos \,\theta }}{{{{\sin }^8}\theta }} $
$= \frac{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}} + \frac{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}}$
$ = \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}}}} \right)$.
