a)$ \frac{tan x}{sin x} $ – $\frac{sinx}{cotx}$ = $cos x$ B) $\frac{1-2sin^{2}x}{1+2sinxcosx}$ = $\frac{1-tanx}{1+tanx}$ c) $cot^{2}x -cos^{2}x$ =

0 bình luận về “a)$ \frac{tan x}{sin x} $ – $\frac{sinx}{cotx}$ = $cos x$ B) $\frac{1-2sin^{2}x}{1+2sinxcosx}$ = $\frac{1-tanx}{1+tanx}$ c) $cot^{2}x -cos^{2}x$ =”

1. $a)\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cot x}\\ =\dfrac{\dfrac{\sin x}{\cos x}}{\sin x}-\dfrac{\sin x}{\dfrac{\cos x}{\sin x}}\\ =\dfrac{1}{\cos x}-\dfrac{\sin^2x}{\cos x}\\ =\dfrac{1-\sin^2x}{\cos x}\\ =\dfrac{\cos^2x}{\cos x}\\ =\cos x\\ b)\dfrac{1-2\sin^2x}{1+2\sin x\cos x}\\ =\dfrac{\dfrac{1}{\cos^2x}-2\tan^2x}{\dfrac{1}{\cos^2x}+2\tan x}\\ =\dfrac{1+\tan^2x-2\tan^2x}{1+\tan^2x+2\tan x}\\ =\dfrac{1-\tan^2x}{(\tan x+1)^2}\\ =\dfrac{(1-\tan x)(\tan x+1)}{(\tan x+1)^2}\\ =\dfrac{1-\tan x}{\tan x+1}\\ c)\cot^2x-\cos^2x\\ =\dfrac{\cos^2x}{\sin^2x}-\dfrac{\cos^2x\sin^2x}{\sin^2x}\\ =\dfrac{\cos^2x-\cos^2x\sin^2x}{\sin^2x}\\ =\dfrac{\cos^2x(1-\sin^2x)}{\sin^2x}\\ =\dfrac{\cos^4x}{\sin^2x}\\ =\dfrac{\cos^2x.\cos^2x}{\sin^2x}\\ =\cot^2x.\cos^2x$  

   Trả lời

2. $a)\quad \dfrac{\tan x}{\sin x} -\dfrac{\sin x}{\cot x}$

   $=\dfrac{\dfrac{\sin x}{\cos x}}{\sin x} -\dfrac{\sin x}{\dfrac{\cos x}{\sin x}}$

   $=\dfrac{1}{\cos x} -\dfrac{\sin^2x}{\cos x}$

   $=\dfrac{1 -\sin^2x}{\cos x}$

   $=\dfrac{\cos^2x}{\cos x}$

   $=\cos x$

   $b)\quad \dfrac{1 -2\sin^2x}{1 + 2\sin x\cos x}$

   $=\dfrac{\cos^2x +\sin^2x -2\sin^2x}{\sin^2x +2\sin x\cos x +\cos^2x}$

   $=\dfrac{\cos^2x -\sin^2x}{(\sin x +\cos x)^2}$

   $=\dfrac{(\cos x -\sin x)(\cos x +\sin x)}{(\sin x +\cos x)^2}$

   $=\dfrac{\cos x -\sin x}{\cos x +\sin x}$

   $=\dfrac{\dfrac{\cos x -\sin x}{\cos x}}{\dfrac{\cos x +\sin x}{\cos x}}$

   $=\dfrac{1 -\tan x}{1 +\tan x}$

   $c)\quad \cot^2x -\cos^2x$

   $=\dfrac{\cos^2x}{\sin^2x} – \cos^2x$

   $=\cos^2x\left(\dfrac{1}{\sin^2x} -1\right)$

   $=\cos^2x\cdot\dfrac{1 -\sin^2x}{\sin^2x}$

   $=\cos^2x\cdot\dfrac{\cos^2x}{\sin^2x}$

   $=\cos^2x.\cot^2x$

   Trả lời