What is (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) equal to?
0
Given:
(tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)
Used formula:
\(\frac{\tan x+\tan y}{1-\tan x\tan y} =\frac{\cot x +\cot y}{\cot x \cot y -1}\)
Calculation:
(tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)
∵ \(\frac{\tan x+\tan y}{1-\tan x\tan y} =\frac{\cot x +\cot y}{\cot x \cot y -1}\)
∴ tan x + tan y = \(\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{\cot x \cot y -1}\)
= \(\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{\cot x \cot y -1}\) × (1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)
= \(-\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{1-\cot x \cot y }\)× (1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)
= - (cot x + cot y)(1 – tan x tan y) + (cot x + cot y)(1 – tan x tan y)
= 0
∴ The value of (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) is 0.
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