Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

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

\(=tan^2x+1+tan^2x=1+2tan^2x\)

quá đúng

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a) A=\cos x-\cos x+\sin ^{2} x \cdot \dfrac{1}{\cos ^{2} x}-\tan ^{2} x=0A=cosx−cosx+sin2x⋅cos2x1​−tan2x=0

b) B=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{1-\cos x+1+\cos x}{(1-\cos x)(1+\cos x)}}-\sqrt{2}B=sinx1​⋅(1−cosx)(1+cosx)1−cosx+1+cosx​​−2​

\begin{aligned}&=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{1-\cos ^{2} x}}-\sqrt{2}=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{\sin ^{2} x}}-\sqrt{2} \\&=\sqrt{2}\left(\dfrac{1}{\sin ^{2} x}-1\right)=\sqrt{2} \cot ^{2} x\end{aligned}​=sinx1​⋅1−cos2x2​​−2​=sinx1​⋅sin2x2​​−2​=2​(sin2x1​−1)=2​cot2x​

a) Ta có: \(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}=1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\)  (Đk: sinx và cosx khác 0)

\(=1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\cos x+\sin x}\)

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

\(=1-\left(\sin^2x+\cos^2x-\sin x.\cos x\right)\) (do sinx + cosx luôn khác 0)

\(=\sin x.\cos x\) ( do \(\sin^2x+\cos^2x=1\))

b) Ta có: \(\frac{\sin^2x+2\cos x-1}{2+\cos x-\cos^2x}=\frac{\left(\sin^2x-1\right)+2\cos x}{-\left(\cos x+1\right)\left(\cos x-2\right)}\) (Đk: cosx khác -1 và 2)

\(=\frac{-\cos x\left(\cos x-2\right)}{-\left(\cos x+1\right)\left(\cos x-2\right)}\)

\(=\frac{\cos x}{1+\cos x}\)

a) Ta có: 1−sin2x1+cotx −cos2x1+tanx =1−sin2x1+cosxsinx  −cos2x1+sinxcosx    (Đk: sinx và cosx khác 0)

=1−sin3xsinx+cosx −cos3xcosx+sinx 

=1−(sinx+cosx)(sin2x−sinx.cosx+cos2x)sinx+cosx 

=1−(sin2x+cos2x−sinx.cosx) (do sinx + cosx luôn khác 0)

=sinx.cosx ( do sin2x+cos2x=1)

b) Ta có: sin2x+2cosx−12+cosx−cos2x =(sin2x−1)+2cosx−(cosx+1)(cosx−2)  (Đk: cosx khác -1 và 2)

=−cosx(cosx−2)−(cosx+1)(cosx−2) 

=cosx1+cosx 

ăerw

a) \(\dfrac{1}{1+tan\alpha}+\dfrac{1}{1+cot\alpha}\)

\(=\dfrac{1}{1+\dfrac{1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)

\(=\dfrac{1}{\dfrac{cot\alpha+1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)

\(=\dfrac{cot\alpha}{cot\alpha+1}+\dfrac{1}{1+cot\alpha}\)

\(=\dfrac{cot\alpha+1}{cot\alpha+1}=1\) (đpcm)

b) \(tan^2x+cot^2x+2\)

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

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

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

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

c) \(sinx.cosx.\left(1+tanx\right)\left(1+cotx\right)\)

\(=\left(sinx.cosx+sinx.cosx.tanx\right)\left(1+cotx\right)\)

\(=\left(sinx.cosx+sinx.cosx.\dfrac{sinx}{cosx}\right)\left(1+cotx\right)\)

\(=\left(sinx.cosx+sin^2x\right)\left(1+cotx\right)\)

\(=\left(sinx.cosx+sin^2x\right)\left(1+\dfrac{cosx}{sinx}\right)\)

\(=sinx.cosx+cos^2x+sin^2x+sinx.cosx\)

\(=1+sin^2x.cos^2x\)

Câu cuối không biết chỗ sai, mong mọi người chỉ bảo ạ ^^

a/ \(\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}\)

\(\Leftrightarrow-2\cos^2x+2\cos x-2\cos x+2\cos^2x=0\)

\(\Leftrightarrow0=0\) (đúng)

\(\RightarrowĐPCM\)

b/ \(\tan a.\tan b=\dfrac{\tan a+\tan b}{\cot a+\cot b}\)

\(\Leftrightarrow\tan a.\tan b.\left(\cot a+\cot b\right)=\tan a+\tan b\)

\(\Leftrightarrow\tan a.\tan b.\cot a+\tan a.\tan b.\cot b=\tan a+\tan b\)

\(\Leftrightarrow\tan b+\tan a=\tan a+\tan b\) (đúng)

\(\RightarrowĐPCM\)

câu 1 : ta có : \(A=\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-\left(sin^8x+cos^8x\right)\)

\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-3sin^2x.cos^2x\right)\)

\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-sin^2x.cos^2x\right)+2sin^2xcos^2x\)

\(=-sin^2x.cos^2x\left(1-sin^2x.cos^2x\right)+2sin^2x.cos^2x\)

\(=sin^2x.cos^2x\left(1+sin^2x.cos^2x\right)\)

tới đây mk xin sử dụng kiến thức lớp 10 một chút

\(=\dfrac{sin^22x}{4}\left(1+\dfrac{sin^22x}{4}\right)=\dfrac{sin^22x}{4}+\dfrac{sin^42x}{16}\)

vẩn phụ thuộc vào x \(\Rightarrow\) đề sai .

câu 1 : câu này bn có thể tìm trong trang của mk , mk nhớ đã làm nó rồi nhưng tìm hoài không đc . nếu đc bn có thể chờ mk đi hok về mk sẽ kiếm cho bn hoắc có thể là lm lại cho bn nha :)

câu 2 : https://hoc24.vn/hoi-dap/question/657072.html

câu 3 : https://hoc24.vn/hoi-dap/question/657069.html

câu 4 : https://hoc24.vn/hoi-dap/question/656635.html

câu 5 : https://hoc24.vn/hoi-dap/question/657071.html

1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)

\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)

\(=1\)

2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)

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

\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)

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

=>VT=VP

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Câu 1 đề vẫn có vấn đề:

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

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)

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

Có thể coi như ko thể rút gọn tiếp

2.

\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

\(=2\left(cos^2x+sin^2x\right)+2=4\)

1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)

\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)

\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )

b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)

\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)

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

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)

\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)

\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )

c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)

\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)

\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)

\(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)

\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)

\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )

d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)

\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )