How do you verify the identity $\frac{tan x + tan y}{1 - tan x tan y} = \frac{cot x + cot y}{cot x cot y - 1}$ $(tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)$ ?

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How do you verify the identity $\frac{tan x + tan y}{1 - tan x tan y} = \frac{cot x + cot y}{cot x cot y - 1}$ $(tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)$ ?

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Answer 1 · 2017-04-19T06:20:13

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Explanation:

Right Hand Side:
$\frac{cot x + cot y}{cot x cot y - 1} = \frac{\frac{1}{tan x} + \frac{1}{tan y}}{\frac{1}{tan x} \cdot \frac{1}{tan y} - 1}$ $color(blue)((cot x+cot y)/(cot xcoty-1)=color(red)((1/tanx + 1/tany)/(1/tanx * 1/tany -1$

$= \frac{\frac{tan y + tan x}{tan x tan y}}{\frac{1 - tan x tan y}{tan x tan y}}$ $color(red)(=((tany+tanx)/(tanxtany))/((1-tanxtany)/(tanx tany)$

$= \frac{tan y + tan x}{tan x tan y} \cdot \frac{tan x tan y}{1 - tan x tan y}$ $color(red)(=(tany+tanx)/(tanxtany) *(tanx tany)/ (1-tanxtany)$

$color(red)(=(tany+tanx)/cancel(tanxtany) *cancel(tanx tany)/ (1-tanxtany)$

$color(red)(=(tany+tanx)/ (1-tanxtany)$

$color(red)(=(tanx+tany)/ (1-tanxtany)$

$color(red)(=$ Left Hand Side

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How do you verify the identity $\frac{tan x + tan y}{1 - tan x tan y} = \frac{cot x + cot y}{cot x cot y - 1}$ $(tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)$ ?

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Explanation:

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How do you verify the identity (tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)tanx+tany1−tanxtany=cotx+cotycotxcoty−1(tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)?

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Explanation:

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How do you verify the identity $\frac{tan x + tan y}{1 - tan x tan y} = \frac{cot x + cot y}{cot x cot y - 1}$ $(tanx+tany)/(1-tanxtany)=(cotx+coty)/(cotxcoty-1)$ ?