How do you prove $\frac{tan x + cot x}{sec x + csc x} = \frac{1}{cos x + sin x}$ $(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)$ ?

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How do you prove $\frac{tan x + cot x}{sec x + csc x} = \frac{1}{cos x + sin x}$ $(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)$ ?

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Answer 1 · 2018-03-09T18:52:35

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Explanation:

$L H S : \frac{tan x + cot x}{sec x + csc x}$ $LHS : (tan x+cot x)/(sec x + csc x)$

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

$= \frac{\frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x}}{\frac{sin x + cos x}{sin x cos x}}$ $=((sin^2x+cos^2x)/(sinxcosx))/((sinx+cosx)/(sinxcosx))$ ->common denominator

$= \frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x} \cdot \frac{sin x cos x}{sin x + cos x}$ $=(sin^2x+cos^2x)/(sinxcosx) *(sinxcosx)/ (sinx+cosx)$

$=(sin^2x+cos^2x)/cancel(sinxcosx) *cancel(sinxcosx)/ (sinx+cosx)$

$=(sin^2x+cos^2x)/(sinx+cosx)$ ->use property $sin^2x+cos^2x=1$

$=1/(sinx+cosx)$

$=RHS$

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How do you prove $\frac{tan x + cot x}{sec x + csc x} = \frac{1}{cos x + sin x}$ $(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)$ ?

1 Answer

Explanation:

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How do you prove (tanx+cotx)/(secx+cscx)=1/(cosx+sinx)tanx+cotxsecx+cscx=1cosx+sinx(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)?

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Explanation:

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How do you prove $\frac{tan x + cot x}{sec x + csc x} = \frac{1}{cos x + sin x}$ $(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)$ ?