3/

\(\frac{sin2x-sinx}{1-cosx+cos2x}=\frac{2sinxcosx-sinx}{1-cosx+2cos^2x-1}=\frac{sinx\left(2cosx-1\right)}{cosx\left(2cosx-1\right)}=\frac{sinx}{cosx}=tanx\)

4/

\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\left(\frac{sinx+\frac{1}{tanx}}{1+sinxtanx}\right)^{2014}=\left(\frac{sinxtanx+1}{tanx\left(sinxtanx+1\right)}\right)^{2014}\)

\(=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)

\(\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}=\frac{sin^{2014}x+\frac{1}{tan^{2014}x}}{1+\left(sinx.tanx\right)^{2014}}=\frac{\left(sinxtanx\right)^{2014}+1}{tan^{2014}x\left[\left(sinxtanx\right)^{2014}+1\right]}\)

\(=\frac{1}{tan^{2014}x}=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)

\(\Rightarrow\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}\)

\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)

=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)

d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)

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

=\(\frac{1}{cosx.sinx}=VP\)

e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)

c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)

=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)

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

Đây nha bạn

Ta có: \(\frac{sinx+cotx}{1+tanx.sinx}=\frac{sinx.cosx\left(sinx+cotx\right)}{sinx.cosx\left(1+tanx.sinx\right)}=\frac{cosx\left(sin^2x+cosx\right)}{sinx\left(cosx+sin^2x\right)}=cotx\)

\(\Rightarrow\frac{\left(sinx+cotx\right)^{2016}}{\left(1+tanx.sinx\right)^{2016}}=cot^{2016}x\) (1)

\(\frac{sin^{2016}x+cot^{2016}x}{1+tan^{2016}x.sin^{2016}x}=\frac{sin^{2016}x.cos^{2016}x\left(sin^{2016}x+cot^{2016}x\right)}{sin^{2016}x.cos^{2016}x\left(1+tan^{2016}x.sin^{2016}x\right)}\)

\(=\frac{cos^{2016}x\left(sin^{4032}x+cos^{2016}x\right)}{sin^{2016}x\left(cos^{2016}x+sin^{4032}x\right)}=cot^{2016}x\) (2)

(1);(2) suy ra đpcm

b)
Với n = 1.
\(VT=B_n=1;VP=\dfrac{1\left(1+1\right)\left(1+2\right)}{6}=1\).
Vậy với n = 1 điều cần chứng minh đúng.
Giả sử nó đúng với n = k.
Nghĩa là: \(B_k=\dfrac{k\left(k+1\right)\left(k+2\right)}{6}\).
Ta sẽ chứng minh nó đúng với \(n=k+1\).
Nghĩa là:
\(B_{k+1}=\dfrac{\left(k+1\right)\left(k+1+1\right)\left(k+1+2\right)}{6}\)\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\).
Thật vậy:
\(B_{k+1}=B_k+\dfrac{\left(k+1\right)\left(k+2\right)}{2}\)\(=\dfrac{k\left(k+1\right)\left(k+2\right)}{6}+\dfrac{\left(k+1\right)\left(k+2\right)}{2}\)\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\).
Vậy điều cần chứng minh đúng với mọi n.

c)
Với \(n=1\)
\(VT=S_n=sinx\); \(VP=\dfrac{sin\dfrac{x}{2}sin\dfrac{2}{2}x}{sin\dfrac{x}{2}}=sinx\)
Vậy điều cần chứng minh đúng với \(n=1\).
Giả sử điều cần chứng minh đúng với \(n=k\).
Nghĩa là: \(S_k=\dfrac{sin\dfrac{kx}{2}sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}\).
Ta cần chứng minh nó đúng với \(n=k+1\):
Nghĩa là: \(S_{k+1}=\dfrac{sin\dfrac{\left(k+1\right)x}{2}sin\dfrac{\left(k+2\right)x}{2}}{sin\dfrac{x}{2}}\).
Thật vậy từ giả thiết quy nạp ta có:
\(S_{k+1}-S_k\)\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}sin\dfrac{\left(k+2\right)x}{2}}{sin\dfrac{x}{2}}-\dfrac{sin\dfrac{kx}{2}sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}\)
\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}.\left[sin\dfrac{\left(k+2\right)x}{2}-sin\dfrac{kx}{2}\right]\)
\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}.2cos\dfrac{\left(k+1\right)x}{2}sim\dfrac{x}{2}\)\(=2sin\dfrac{\left(k+1\right)x}{2}cos\dfrac{\left(k+1\right)x}{2}=2sin\left(k+1\right)x\).
Vì vậy \(S_{k+1}=S_k+sin\left(k+1\right)x\).
Vậy điều cần chứng minh đúng với mọi n.

iểm tra với n = 1

Giả sử đã có 

Viết S k + 1   =   S k   +   sin ( k + 1 ) x sử dụng giả thiết quy nạp và biến đổi ta có

\(\frac{sin^2a+1}{2.cos^2a}+\frac{1+cos^2a}{2.sin^2a}+1=\frac{tan^2a}{2}+\frac{1}{2cos^2a}+\frac{cot^2a}{2}+\frac{1}{2sin^2a}+1\)

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

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

B.

\(\frac{1-4sin^2a.cos^2a}{4sin^2a.cos^2a}=\frac{\frac{1}{cos^4a}-\frac{4sin^2a}{cos^2a}}{\frac{4sin^2a}{cos^2a}}=\frac{\left(\frac{1}{cos^2a}\right)^2-4tan^2a}{4tan^2a}=\frac{\left(1+tan^2a\right)^2-4tan^2a}{4tan^2a}\)

\(=\frac{tan^4a-2tan^2a+1}{4tan^2a}\)

C.

\(\frac{sina+tana}{tana}=\frac{sina}{tana}+1=1+sina.\frac{cosa}{sina}=1+cosa\)

D.

\(tana+\frac{cosa}{1+sina}=\frac{sina}{cosa}+\frac{cosa\left(1-sina\right)}{1-sin^2a}=\frac{sina.cosa}{cos^2a}+\frac{cosa-cosa.sina}{cos^2a}\)

\(=\frac{sina.cosa+cosa-sina.cosa}{cos^2a}=\frac{cosa}{cos^2a}=\frac{1}{cosa}\)

Câu C sai