Answer 1.2:1

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a) \displaystyle \cos^2x-\sin^2x=\cos2x b) \displaystyle 2x\ln x+ x c) \displaystyle \displaystyle\frac{x^2+2x-1}{(x+1)^2}=1-\frac{2}{(x+1)^2}
d) \displaystyle \displaystyle\frac{\cos x}{x}-\frac{\sin x}{x^2} e) \displaystyle \displaystyle\frac{1}{\ln x}-\frac{1}{(\ln x)^2} f) \displaystyle \displaystyle \frac{\ln x + 1}{\sin x}-\frac{x\ln x \cos x}{\sin^2x}