tan x + cot x DIVIDE by csc x ?
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Please clarify: Do you mean (tan x + cot x)/(csc x)
or tan x + (cot x)/(csc x) ?
The first one is (sin x)(tan x) + (cos x)
The second one is tan x + cos x
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You still didn't clarify the equation. Let's assume you meant the first one I wrote.
Note that tan x = (sin x)/(cos x), cot x = (cos x)/(sin x), and csc x = 1/(sin x)
(tan x + cot x)/(csc x) = [(sin x)/(cos x) + (cos x)/(sin x)] × (sin x)
= (sin x)²/(cos x) + (cos x)
= (sin x)²/(cos x) + (cos x)²/(cos x)
= [(sin x)² + (cos x)²]/(cos x) ... here comes the fun part!
= 1/(cos x)
= sec x
or tan x + (cot x)/(csc x) ?
The first one is (sin x)(tan x) + (cos x)
The second one is tan x + cos x
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You still didn't clarify the equation. Let's assume you meant the first one I wrote.
Note that tan x = (sin x)/(cos x), cot x = (cos x)/(sin x), and csc x = 1/(sin x)
(tan x + cot x)/(csc x) = [(sin x)/(cos x) + (cos x)/(sin x)] × (sin x)
= (sin x)²/(cos x) + (cos x)
= (sin x)²/(cos x) + (cos x)²/(cos x)
= [(sin x)² + (cos x)²]/(cos x) ... here comes the fun part!
= 1/(cos x)
= sec x
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That's the same as (tan(x)+cot(x))*sin(x)=(sin^2(x))/cos(x)… Put them both over cosine and you get (sin^2(x)+cos^2(x))/cos(x)=1/cos(x)=sec(…