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

Question

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

$c)D$=$\frac{cot^{2}x-cos^{2}x}{cos^{2}x}$ $+1-Cos^{2}x$
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Isabelle 7 phút 2021-09-09T17:48:21+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-09T17:49:29+00:00

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

     

    0
    2021-09-09T17:49:40+00:00

    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,

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

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

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

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

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

    $=\cot^2x+\sin^2x$ 

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