a) Ta có:

      \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)

b) Ta có:

\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

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

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

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

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)

\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)

\(=sinx+cosx-cosx=sinx\)

\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)

\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)

\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)

\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)

\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)

Đúng như bạn viết vế trái là thế này:

\(\left(\frac{tan^2x}{1+tan^2x}\right)\left(\frac{1+cot^2x}{cotx}\right)=\left(\frac{1}{\frac{1}{tan^2x}+1}\right)\left(\frac{1+cot^2x}{cotx}\right)\)

\(=\left(\frac{1}{cot^2x+1}\right)\left(\frac{1+cot^2x}{cotx}\right)=\frac{1}{cotx}=tanx\)

Còn vế phải sẽ ra thế này:

\(\frac{1+tan^4x}{tan^2x+cot^2x}=\frac{1+tan^4x}{tan^2x+\frac{1}{tan^2x}}=\frac{tan^2x\left(1+tan^4x\right)}{tan^4x+1}=tan^2x\)

Hai vế ra kết quả khác nhau nên chắc bạn ghi sai đề :)

\(a,VT=2.tanx+tan\left(-x\right)\\ =2tanx-tanx=tanx\)

\(b,VT=sin\left(2\pi+\dfrac{\pi}{2}-x\right)+cos\left(12\pi+\pi+x\right)-sin\left(x-4\pi-\pi\right)\\ =sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi+x\right)+sin\left(\pi-x\right)\\ =cosx-cosx+sinx\\ =sinx=VP\)

a: \(2\cdot cot\left(\dfrac{pi}{2}-x\right)+tan\left(pi-x\right)\)

\(=2\cdot tanx-tanx\)

=tan x

b: \(sin\left(\dfrac{5}{2}pi-x\right)+cos\left(13pi+x\right)-sin\left(x-5pi\right)\)

\(=sin\left(\dfrac{pi}{2}-x\right)+cos\left(pi+x\right)+sin\left(pi-x\right)\)

\(=cosx-cosx+sinx=sinx\)