Cathy pasted 5 leaves on each picture she made.

Ответ:

lks715

16.11.2021

$\left(1+\cot y-\csc y\right)\left(1+\tan y +\sec y\right) = 2$ i did this two ways: long expansion and difference of squares manipulation.long expansion is first; scroll down if you want to see the other method.it is messy, but we can expand out the right side.we expand it like this, like what you would do with multiplying polynomials: $(a+b-c)(a+d+e)\\ = a(a+d+e) + b(a+d+e) - c(a+d+e)$ so $(1+\cot y-\csc y)(1+\tan y +\sec y)$ $\begin{multiline} 1(1+\tan y +\sec y) + \cot y (1+\tan y +\sec y) \\ \phantom{=} - \csc y(1+\tan y +\sec y) \end{multiline}$ $\begin{multiline} 1+\tan y +\sec y + \cot y+ \cot y\tan y +\cot y\sec y \\ \phantom{=} - \csc y - \csc y\tan y - \csc y\sec y \end{multiline}$ we turn everything into sines and cosines. since $\begin{aligned} \tan y & = \tfrac{\sin y}{\cos y} & \sec y & = \tfrac{1}{\cos y} \\ \cot y & = \tfrac{\cos y}{\sin y} & \csc y & = \tfrac{1}{\sin y} \end{aligned}$ then $\begin{multiline} 1+\tfrac{\sin y}{\cos y} + \tfrac{1}{\cos y} + \tfrac{\cos y}{\sin y}+ \tfrac{\cos y}{\sin y}\tfrac{\sin y}{\cos y} +\tfrac{\cos y}{\sin y}\tfrac{1}{\cos y} \\ \phantom{=} - \tfrac{1}{\sin y} - \tfrac{1}{\sin y} \tfrac{\sin y}{\cos y} - \tfrac{1}{\sin y} \tfrac{1}{\cos y} \end{multiline}$ some expressions can be canceled, e.g. $\begin{aligned} \tfrac{\cos y}{\sin y}\tfrac{\sin y}{\cos y} & \rightarrow 1 \\ \tfrac{\cos y}{\sin y}\tfrac{1}{\cos y} & \rightarrow \tfrac{1}{\sin y} \\ \tfrac{1}{\sin y} \tfrac{\sin y}{\cos y} & \rightarrow \tfrac{1}{\cos y} \end{aligned}$ so $\begin{multiline} 1+\tfrac{\sin y}{\cos y} + \tfrac{1}{\cos y} + \tfrac{\cos y}{\sin y}+ 1+\tfrac{1 }{\sin y} \\ \phantom{=} - \tfrac{1}{\sin y} - \tfrac{1}{\cos y} - \tfrac{1}{\sin y\cos y} \end{multiline}$ we have like terms: $+ \tfrac{1}{\cos y} \text{ and } - \tfrac{1}{\cos y}$ $+ \tfrac{1}{\sin y} \text{ and } - \tfrac{1}{\sin y}$ they add with each other to become zero, so doing that $1+\tfrac{\sin y}{\cos y} + \tfrac{\cos y}{\sin y}+ 1 - \tfrac{1}{\sin y\cos y}$ 1 and 1 add to become 2, so doing that $2+\tfrac{\sin y}{\cos y} + \tfrac{\cos y}{\sin y}- \tfrac{1}{\sin y\cos y}$ we want to have everything have a common denominator of $\sin y\cos y$ multiply numerator and denominator of 2 by $\sin y\cos y$ $\tfrac{2\sin y\cos y}{\sin y\cos y}+\tfrac{\sin y}{\cos y} + \tfrac{\cos y}{\sin y}- \tfrac{1}{\sin y\cos y}$ multiply numer. and denom. of $\tfrac{\sin y}{\cos y}$ by $\sin y$ $\tfrac{2\sin y\cos y}{\sin y\cos y}+\tfrac{\sin^2 y}{\sin y\cos y} + \tfrac{\cos y}{\sin y}- \tfrac{1}{\sin y\cos y}$ multiply numer. and denom. of $\tfrac{\cos y}{\sin y}$ by $\cos y$ $\tfrac{2\sin y\cos y}{\sin y\cos y}+\tfrac{\sin^2 y}{\sin y\cos y} + \tfrac{\cos^2 y}{\sin y\cos y}- \tfrac{1}{\sin y\cos y}$ they all have like denominators, so combining $\dfrac{2\sin y\cos y + \sin^2 y + \cos^2 y - 1}{\sin y\cos y}$ the pythagorean identity says that $\sin^2 y + \cos^2 y = 1$ so we can substitute 1 in for $\sin^2 y + \cos^2 y$ $\dfrac{2\sin y\cos y + 1 - 1}{\sin y\cos y}$ 1 - 1 is 0 $\dfrac{2\sin y\cos y}{\sin y\cos y}$ $\sin y\cos y$ cancel and you are left with $2$ which is the right-hand side.======alternatively $\begin{aligned} \text{lhs} & = \left(1+\cot y-\csc y\right)\left(1+\tan y +\sec y\right) \\ & = \left(1+\tfrac{\cos y}{\sin y}-\tfrac{1}{\sin y}\right)\left(1+\tfrac{\sin y}{\cos y} +\tfrac{1}{\cos y}\right) \\ & = \left(\tfrac{\sin y}{\sin y}+\tfrac{\cos y}{\sin y}-\tfrac{1}{\sin y}\right)\left(\tfrac{\cos y}{\cos y}+\tfrac{\sin y}{\cos y} +\tfrac{1}{\cos y}\right) \end{aligned}$ combine $\begin{aligned} \text{lhs} & = \left(\frac{\sin y + \cos y - 1}{\sin y}\right)\left(\frac{\cos y + \sin y + 1}{\cos y}\right) \\ & = \left(\frac{\sin y + \cos y - 1}{\sin y}\right)\left(\frac{\sin y + \cos y + 1}{\cos y}\right) \end{aligned}$ let $a = (\sin y + \cos y)$ $\begin{aligned} \text{lhs} & = \left(\frac{a - 1}{\sin y}\right)\left(\frac{a + 1}{\cos y}\right)\\ & = \frac{(a - 1)(a + 1)}{\sin y\cos y} \\ & = \frac{a^2 - 1}{\sin y \cos y} \\ & = \frac{(\sin y + \cos y)^2 - 1}{\sin y \cos y}\end{aligned}$ $\begin{aligned} (\sin y + \cos y)^2 & \rightarrow \sin^2y + 2\sin y\cos y + \cos^2 y \\ & \rightarrow \boxed{\sin^2y + \cos^2 y} + 2\sin y\cos y \\ & \rightarrow \boxed{1} + 2\sin y \cos y \end{aligned}$ so $\begin{aligned} \text{lhs} & = \frac{(\sin y + \cos y)^2 - 1}{\sin y \cos y}\\ & = \frac{1+ 2\sin y \cos y - 1}{\sin y \cos y} \\ & = \frac{2\sin y \cos y}{\sin y \cos y} \\ & = 2 \\ & = \text{rhs} \end{aligned}$

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Cathy pasted 5 leaves on each picture she made. She used 10 leaves in all. How many pictures did cathy make?

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