$a) A=$$\frac{(sinx+cox)^{2}-1}{cotx-sinxcosx}$ $b) B$=$\frac{sin^{2}x+cos^{2}x-1}{cot^{2}x}$ $c)C$=$\frac{sin^{2}x-tan^{2}x}{cos^{2}x-cot^{2}x}$

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

1. \(\begin{array}{l}
   a)\\
   \quad A = \dfrac{(\sin x + \cos x)^2 – 1}{\cot x – \sin x\cos x}\\
   \to A = \dfrac{\sin^2x + \cos^2x + 2\sin x\cos x – 1}{\dfrac{\cos x}{\sin x} – \sin x\cos x}\\
   \to A = \dfrac{1 + 2\sin x\cos x – 1}{\dfrac{\cos x – \sin^2x\cos x}{\sin x}}\\
   \to A = \dfrac{2\sin x\cos x}{\cos x\cdot\dfrac{1- \sin^2x}{\sin x}}\\
   \to A = \dfrac{2\sin^2x}{1 – \sin^2x}\\
   \to A = \dfrac{2\sin^2x}{\cos^2x}\\
   \to A = 2\tan^2x\\
   b)\\
   \quad B = \dfrac{\sin^2x + \cos^2x – 1}{\cot^2x}\\
   \to B = \dfrac{1-1}{\cot^2x}\\
   \to B = \dfrac{0}{\cot^2x}\\
   \to B = 0\\
   c)\\
   \quad C = \dfrac{\sin^2x – \tan^2}{\cos^2x – \cot^2x}\\
   \to C = \dfrac{\sin^2x – \dfrac{\sin^2x}{\cos^2x}}{\cos^2x – \dfrac{\cos^2x}{\sin^2x}}\\
   \to C = \dfrac{\sin^2x\left(1 – \dfrac{1}{\cos^2x}\right)}{\cos^2x\left(1 – \dfrac{1}{\sin^2x}\right)}\\
   \to C = \dfrac{\sin^2x\cdot \dfrac{\cos^2x – 1}{\cos^2x}}{\cos^2x\cdot \dfrac{\sin^2x – 1}{\sin^2x}}\\
   \to C = \dfrac{-\sin^4x(1 – \cos^2x)}{-\cos^4x(1 – \sin^2x)}\\
   \to C = \dfrac{\sin^4x.\sin^2x}{\cos^4x.\cos^2x}\\
   \to C = \dfrac{\sin^6x}{\cos^6x}\\
   \to C = \tan^6x\\
   d)\\
   \quad D = \dfrac{\cot^2x – \cos^2x}{\cos^2x}  + 1 – \cot^2x\\
   \to D = \dfrac{\cot^2x}{\cos^2x} – 1 + 1 – \cot^2x\\
   \to D = \dfrac{\dfrac{\cos^2x}{\sin^2x}}{\cos^2x} – \cot^2x\\
   \to D = \dfrac{1}{\sin^2x} – \cot^2x\\
   \to D = \dfrac{\sin^2x + \cos^2x}{\sin^2x} – \cot^2x\\
   \to D ​= 1 + \dfrac{\cos^2x}{\sin^2x} – \cot^2x\\
   \to D = 1 + \cot^2x – \cot^2x\\
   \to D ​= 1
   \end{array}\)

    

   Trả lời

2. a,

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

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

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

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

   $=\dfrac{2\sin^2x\cos x}{\cos^3x}$

   $=2\tan^2x$   

   b,

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

   $=\dfrac{1-1}{\cot^2x}$

   $=0$

   c,

   $C=\dfrac{\sin^2x-\tan^2x}{\cos^2x-\cot^2x}$

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

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

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

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

   $=\tan^6x$

   d,

   $D=\dfrac{\cot^2x-\cos^2x}{\cos^2x}+1-\cos^2x$

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

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

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

   $=1+\cot^2x-\cos^2x$

   $=\cot^2x+\sin^2x$ 

   Trả lời