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Slav-nsk [51]
3 years ago
6

Simplify (1-cos theta)(1+cos theta) over (1-sin theta) (1+ sin theta)

Mathematics
1 answer:
slega [8]3 years ago
4 0
\bf sin^2(\theta)+cos^2(\theta)=1\to 
\begin{cases}
cos^2(\theta)=1-sin^2(\theta)\\
sin^2(\theta)=1-cos^2(\theta)
\end{cases} \\\\
-------------------------------\\\\
\stackrel{\textit{difference of squares}}{\cfrac{[1-cos(\theta )][1+cos(\theta )]}{[1-sin(\theta )][1+sin(\theta )]}}\implies \cfrac{1^2-cos^2(\theta )}{1^2-sin^2(\theta )}\implies \cfrac{1-cos^2(\theta )}{1-sin^2(\theta )}
\\\\\\
\cfrac{sin^2(\theta )}{cos^2(\theta )}\implies tan^2(\theta )
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\gray{ \frak{The \:  given \:  two \:   \: points  \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}

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Let's solve by using midpoint formula :

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\bf \boxed{\color{red}\frak{Midpoint \:   \: Formula :  {( \: x ,\:y \:  ) = }(  \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}

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\: \large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{  \cancel\frac{ -4}{2} ,  \cancel\frac{ - 5}{2} )}}

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\underline{ \boxed{ \large \red{ \frak{option \: d }(  \frak{   - 2, - 2.5)}}}}✓

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