Chứng minh hệ thức không phụ thuộc vào x B = (1 + cotx) / (1-cotx) – (2 +2cot ²x)/(Tanx -1)(tan ²x +1) 03/09/2021 Bởi Raelynn Chứng minh hệ thức không phụ thuộc vào x B = (1 + cotx) / (1-cotx) – (2 +2cot ²x)/(Tanx -1)(tan ²x +1)
Giải thích các bước giải: Ta có: $B=\dfrac{1+\cot x}{1-\cot x}-\dfrac{2+2\cot^2x}{(\tan x-1)(\tan^2x+1)}$ $\to B=\dfrac{\tan x\cot x+\cot x}{\tan x\cot x-\cot x}-\dfrac{2(1+\cot^2x)}{(\tan x-1)(\tan^2x+1)}$ $\to B=\dfrac{\cot x(\tan x+1)}{\cot x(\tan x-1)}-\dfrac{2(\cot^2x\tan^2x+\cot^2x)}{(\tan x-1)(\tan^2x+1)}$ $\to B=\dfrac{\tan x+1}{\tan x-1}-\dfrac{2\cot^2x(\tan^2x+1)}{(\tan x-1)(\tan^2x+1)}$ $\to B=\dfrac{\tan x+1}{\tan x-1}-\dfrac{2\cot^2x}{\tan x-1}$ $\to B=\dfrac{\tan x+1-2\cot^2x}{\tan x-1}$ $\to B=\dfrac{\tan x-1+2-2\cot^2x}{\tan x-1}$ $\to B=1+\dfrac{2-2\cot^2x}{\tan x-1}$ $\to B=1+\dfrac{2(1-\cot^2x)}{\tan x-1}$ $\to B=1+\dfrac{2(1-\cot x)(1+\cot x)}{\tan x-1}$ $\to B=1+\dfrac{2(\tan x\cot x-\cot x)(1+\cot x)}{\tan x-1}$ $\to B=1+\dfrac{2\cot x(\tan x-1)(1+\cot x)}{\tan x-1}$ $\to B=1+2\cot x(1+\cot x)$ $\to B=1+2\cot x+2\cot^2x$ Bình luận
Giải thích các bước giải:
Ta có:
$B=\dfrac{1+\cot x}{1-\cot x}-\dfrac{2+2\cot^2x}{(\tan x-1)(\tan^2x+1)}$
$\to B=\dfrac{\tan x\cot x+\cot x}{\tan x\cot x-\cot x}-\dfrac{2(1+\cot^2x)}{(\tan x-1)(\tan^2x+1)}$
$\to B=\dfrac{\cot x(\tan x+1)}{\cot x(\tan x-1)}-\dfrac{2(\cot^2x\tan^2x+\cot^2x)}{(\tan x-1)(\tan^2x+1)}$
$\to B=\dfrac{\tan x+1}{\tan x-1}-\dfrac{2\cot^2x(\tan^2x+1)}{(\tan x-1)(\tan^2x+1)}$
$\to B=\dfrac{\tan x+1}{\tan x-1}-\dfrac{2\cot^2x}{\tan x-1}$
$\to B=\dfrac{\tan x+1-2\cot^2x}{\tan x-1}$
$\to B=\dfrac{\tan x-1+2-2\cot^2x}{\tan x-1}$
$\to B=1+\dfrac{2-2\cot^2x}{\tan x-1}$
$\to B=1+\dfrac{2(1-\cot^2x)}{\tan x-1}$
$\to B=1+\dfrac{2(1-\cot x)(1+\cot x)}{\tan x-1}$
$\to B=1+\dfrac{2(\tan x\cot x-\cot x)(1+\cot x)}{\tan x-1}$
$\to B=1+\dfrac{2\cot x(\tan x-1)(1+\cot x)}{\tan x-1}$
$\to B=1+2\cot x(1+\cot x)$
$\to B=1+2\cot x+2\cot^2x$