Math, asked by anjurockybaby, 11 months ago

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

Answers

Answered by MaheswariS
0

\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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