Question

What is the mid point of (sec^2 x,cot^2 x) and (tan^2 x,-csc^2 x)

Answer 1

$\textbf{Midpoint formula:}$

$\text{The midpoint of the line joining (x_1,y_1) and (x_2,y_2) is}$

$\boxed{(\bf\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})}$

$\textbf{Given points are}$

$(sec^2 x,cot^2 x)\;\text{and}\;(-tan^2 x,-cosec^2 x)$

$\textbf{To find:}$

$\text{The midpoint of the given two points}$

$\text{By using, Midpoint formula,}$

$\text{The midpoint of the given two points is}$

$(\bf\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

$=(\dfrac{sec^2x+(-tan^2x)}{2},\dfrac{cot^2x+(-cosec^2 x)}{2})$

$=(\dfrac{sec^2x-tan^2x}{2},\dfrac{cot^2x-cosec^2 x}{2})$

$=(\dfrac{sec^2x-tan^2x}{2},\dfrac{-(cosec^2x-cot^2 x)}{2})$

$\text{Using the following identities,}$

$\boxed{\bf\,sec^2A-tan^2A=1\;\text{and}\;cosec^2A-cot^2A=1}$

$=(\dfrac{1}{2},\dfrac{-1}{2})$

$\therefore\textbf{The midpoint of the line joining \bf(sec^2 x,cot^2 x) and \bf(-tan^2 x,-cosec^2 x) is (\dfrac{1}{2},\dfrac{-1}{2}) }$

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