Sao có \(cosb\) ở đây??

\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\)

\(=\tan^2\alpha\cdot\left(1-\cos^2\alpha\right)\)

\(=\tan^2\alpha\cdot\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)

\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\\ =\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha\cdot\cos^2\alpha\\ =\dfrac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha=\sin^2\alpha\\ =1-\cos^2\alpha=\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)

\(VT=\dfrac{1+\cos^2a-\sin^2a+2\cdot\sin a\cdot\cos a}{1+2\cdot\sin a\cdot\cos a-\cos^2a+\sin^2a}\)

\(=\dfrac{2\cdot\cos^2a+2\cdot\sin a\cdot\cos a}{2\cdot\sin^2a+2\cdot\sin a\cdot\cos a}\)

\(=\dfrac{2\cdot\cos a\left(\cos a+\sin a\right)}{2\cdot\sin a\cdot\left(\sin a+\cos a\right)}\)

\(=\dfrac{\cos a}{\sin a}=\cot a\)

\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)

\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)

Lời giải:

\(B=\frac{3}{x-1}\sqrt{\frac{(x-1)^2}{(3x)^2}}=\frac{3}{x-1}|\frac{x-1}{3x}|\)

\(=\frac{3}{x-1}.\frac{1-x}{3x}=\frac{-1}{x}\)

\(B=\dfrac{3}{x-1}.\sqrt{\dfrac{x^2-2x+1}{9x^2}}=\dfrac{3}{x-1}.\sqrt{\left(\dfrac{x-1}{3x}\right)^2}\)

\(=\dfrac{3}{x-1}.\left|\dfrac{x-1}{3x}\right|=\dfrac{3}{x-1}.\dfrac{1-x}{3x}=-\dfrac{1}{x}\)

...

đk : x >= 0 ; x khác 1 

\(B=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

B xác định \(< =>\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\left(luôn.đúng\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)