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Andrew [12]
3 years ago
12

Please help, very confused!

Mathematics
1 answer:
My name is Ann [436]3 years ago
7 0
\bf cot(\theta)=\cfrac{1}{tan(\theta)}
\qquad \qquad 
csc(\theta)=\cfrac{1}{sin(\theta)}\qquad \qquad sin(-\theta )=-sin(\theta )
\\\\\\
\textit{also recall }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------

\bf \cfrac{csc^2(x)-cot^2(x)}{sin(-x)cot(x)}\implies \cfrac{\frac{1}{sin^2(x)}-\frac{cos^2(x)}{sin^2(x)}}{-sin(x)\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{1-cos^2(x)}{sin^2(x)}}{-cos(x)}
\\\\\\
\cfrac{\frac{sin^2(x)}{sin^2(x)}}{-cos(x)}\implies \cfrac{1}{-cos(x)}\implies -sec(x)
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n( WuR)=?

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Step-by-step explanation:

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This ratio of side lengths is constant for all corresponding sides.

Therefore, corresponding sides are proportional.

Since, all angles of both the squares are congruent and all the sides are proportional, both the squares will be similar.

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<h3>Answer:</h3>

[B] O.

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