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यदि $\cot \,\theta + \tan \theta = m$ तथा $\sec \theta - \cos \theta = n,$ तब निम्नलिखित में से कौन सा सही है
$m{(m{n^2})^{1/3}} - n{(n{m^2})^{1/3}} = 1$
$m{({m^2}n)^{1/3}} - n{(m{n^2})^{1/3}} = 1$
$n{(m{n^2})^{1/3}} - m{(n{m^2})^{1/3}} = 1$
$n{({m^2}n)^{1/3}} - m{(m{n^2})^{1/3}} = 1$
Solution
दिये गये अनुसार,
$\frac{1}{{\tan \theta }} + \tan \theta = m\,$
$\Rightarrow \,1 + {\tan ^2}\theta = m\,\tan \theta $
$ \Rightarrow \,\,{\sec ^2}\theta = m\,\tan \theta $…..$(i)$
व $\sec \theta – \cos \theta = n\,\, $
$\Rightarrow \,\,{\sec ^2}\theta – 1 = n\,\sec \theta $
$ \Rightarrow \,\,{\tan ^2}\theta = n\,\,\sec \theta $
$ \Rightarrow \,\,{\tan ^4}\theta = {n^2}\,{\sec ^2}\theta = {n^2}.\,m\,\,\tan \theta $ {$(i)$ द्वारा}
$ \Rightarrow \,\,\tan \theta = {({n^2}m)^{1/3}}$…..$(ii)$
तथा ${\sec ^2}\theta = m\,\,\tan \theta = m\,{({n^2}m)^{1/3}}$ {$(i)$ व $(ii)$ से}
$\therefore$ सर्वसमिका ${\sec ^2}\theta – {\tan ^2}\theta = 1$ से,
$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} – {({n^2}m)^{2/3}} = 1$
$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} – n\,{(n{m^2})^{1/3}} = 1.$