InterviewSolution
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Find the Minimum Value of \({\bf{sec}}{\;^2}{\bf{x}}\; + \;{\bf{cosec}}{\;^2}{\bf{x}}\)1). 22). 43). 64). 8 |
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Answer» $({\bf{sec}}{^2}{\bf{x}} + {\bf{cosec}}{^2}{\bf{x}} = 1{\rm{}} + {\rm{}}{\tan ^2}{\rm{x}} + {\rm{}}1{\rm{}} + {\rm{}}{\COT ^2}{\rm{x}} = {\rm{}}2{\rm{}} + {\rm{}}{\tan ^2}{\rm{x}} + {\rm{}}{\cot ^2}{\rm{x}})$ Now we have to find the minimum value of $(\begin{array}{l} \RIGHTARROW {\bf{ta}}{{\bf{n}}^2}{\bf{x}} + {\cot ^2}{\bf{x}} = {\LEFT( {\tan {\bf{x}}} \RIGHT)^2} + {\left( {\cot {\bf{x}}} \right)^2} - 2 \times \tan {\bf{x}} \times \\cot {\bf{x}} + 2 \times \tan {\bf{x}} \times \cot {\bf{x}}\\ \Rightarrow {\left( {\tan {\bf{x}} - \cot {\bf{x}}} \right)^2} + 2 = {\left( {\tan {\bf{x}} - \cot {\bf{x}}} \right)^2} + 2\end{array})$ For the value to be minimum, $(\tan {\bf{x}} - \cot {\bf{x}}\; = \;0)$ ⇒ Minimum value = 2 + 2 = 4 |
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