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Nonamiya [84]
3 years ago
8

Pre-Calculus Help? Polar and Parametric Equations?

Mathematics
1 answer:
Paraphin [41]3 years ago
7 0
\bf x=rcos(\theta )\implies \cfrac{x}{cos(\theta )}=r\\\\
y=rsin(\theta )\implies \cfrac{y}{sin(\theta )}=r
\\\\
-------------------------------\\\\
r=3cos(\theta )\implies \cfrac{x}{cos(\theta )}=3cos(\theta )\implies x=3cos(\theta )cos(\theta )
\\\\\\
\boxed{x=3cos^2(\theta )}
\\\\\\
r=3cos(\theta )\implies \cfrac{y}{sin(\theta )}=3cos(\theta )\implies \boxed{y=3cos(\theta )sin(\theta )}
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prove the following identity: sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x please provide a proof in some shape form or fash
wolverine [178]

Answer:

Step-by-step explanation:

Hello,

<u><em>Is this equality true ?</em></u>

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

<u>1. let 's estimate the left part of the equation</u>

sec(x)csc(x)(tan(x) + cot(x)) =\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin(x)}{cos(x)}+\dfrac{cos(x)}{sin(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin^2(x)+cos^2(x)}{sin(x)cos(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{1}{sin(x)cos(x)})\\\\\\=\dfrac{1}{cos^2(x)sin^2(x)}

<u>1. let 's estimate the right part of the equation</u>

<u />2+tan^2(x) + cot^2(x)=2+\dfrac{sin^2(x)}{cos^2(x)}+\dfrac{cos^2(x)}{sin^2(x)}\\\\=\dfrac{2cos^2(x)sin^2(x)+cos^4(x)+sin^4(x)}{cos^2(x)sin^2(x)}\\\\=\dfrac{(cos^2(x)+sin^2(x))^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1}{cos^2(x)sin^2(x)}<u />

This is the same expression

So

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

hope this helps

7 0
3 years ago
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