Question

Prove that cot(x-y)=cotx×coty+1/ coty-cotx

Answer 1

Answer:

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Answer 2

Given:

cot(x-y)=cotx×coty+1/ coty-cotx

To find:

To prove cot(x-y)=cotx×coty+1/ coty-cotx

Solution:

consider,

$L.H.S = \cot x-y$

As we know the trigonometric identity of $\cot x-y$ is  $\frac{1+\tan x \cdot \tan y}{\tan x-\tan y}$

So,

         $=\frac{1+\tan x \tan y}{\tan x-\tan y}$

Divide the numerator and denominator with $\tan x \cdot \tan y$  

         $=\frac{\frac{1}{\tan x \cdot \tan y}+\frac{\tan x \tan y}{\tan x \cdot \tan y}}{\frac{\tan x-\tan y}{\tan x \cdot \tan y}}$

         $\frac{\cot x \cdot \cot y+1}{\frac{1}{\tan y}-\frac{1}{\tan x}}$

          $= \frac{\cot x \cot y+1}{\cot y-\cot x}$

          $=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$

Hence proved cot(x-y)=cotx×coty+1/ coty-cotx

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