Question

if none of the angles x,y and. (x+y) is an odd multiple of π/2,then prove that tan (x+y)=tanx+tany/1-tanxtany and tan (x-y=tanx-tany/1+tanxtany​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

It is proven that  $tan(x+y) = \frac{tan x+ tany}{1-tanx tany}$  and  $tan(x -y) = \frac{tan x- tan y}{1+tanx tan y}$

Given: x and y are angles  

Tofind: prove that  $tan(x+y) = \frac{tan x+ tany}{1-tanx tany}$  and $tan (x-y) = \frac{tan x - tany}{1+tanx tany}$  

Solution:    

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

Take LHS   $tan(x+y)$

⇒ $tan(x+y) = \frac{sin (x+y)}{cos(x+y)}$      

⇒  $\frac{sinx cosy +cox siny)}{cos x cos y - sinx sin y}$    

⇒ $\frac{sinx cosy/cos x cosy +cox siny/cos x cos y)}{cos x cos y/cos x coy - sinx sin y/cos x cos y}$  

⇒  $\frac{ \frac{sinx}{cos x} + \frac{siny}{cos y} )}{1 - \frac{sinx sin y}{cos x cos y} }$   = $\frac{ tanx + tany)}{1 - tan x tany} }$  = RHS

Hence it is proven that  $tan(x+y) = \frac{tan x+ tany}{1-tanx tany}$  

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

⇒  Here we have  $tan(x+y) = \frac{tan x+ tany}{1-tanx tany}$  take -y instead of y

⇒  $tan(x+(-y)) = \frac{tan x + tan(-y)}{1-tanx tan(-y)}$

⇒  $tan(x -y) = \frac{tan x- tan y}{1+tanx tan y}$        

Hence it is proven that  $tan(x -y) = \frac{tan x- tan y}{1+tanx tan y}$

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