How do you prove $tan x + cot x = sec x csc x$ $Tan x+Cot x =Sec x Csc x$ ?

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How do you prove $tan x + cot x = sec x csc x$ $Tan x+Cot x =Sec x Csc x$ ?

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Answer 1 · 2018-06-11T00:15:53

We have:

$\frac{sin x}{cos x} + \frac{cos x}{sin x} = (\frac{1}{cos x}) (\frac{1}{sin x})$ $sinx/cosx + cosx/sinx = (1/cosx)(1/sinx)$

$\frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x} = \frac{1}{sin x cos x}$ $(sin^2x+ cos^2x)/(sinxcosx) = 1/(sinxcosx)$

Recall that ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x =1$ .

$\frac{1}{sin x cos x} = \frac{1}{sin x cos x}$ $1/(sinxcosx) = 1/(sinxcosx)$

Hopefully this helps!

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How do you prove $tan x + cot x = sec x csc x$ $Tan x+Cot x =Sec x Csc x$ ?

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How do you prove $tan x + cot x = sec x csc x$ $Tan x+Cot x =Sec x Csc x$ ?