\(sina\sqrt{1+\frac{sin^2a}{cos^2a}}=sina\sqrt{\frac{cos^2a+sin^2a}{cos^2a}}=\frac{sina}{\left|cosa\right|}=\pm tana\)

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

\(\frac{1-4sin^2xcos^2x}{\left(sinx+cosx\right)^2}=\frac{\left(1-2sinx.cosx\right)\left(1+2sinx.cosx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(1-sin2x\right)\left(1+2sinx.cosx\right)}{1+2sinx.cosx}=1-2sinx\)

\(sin\left(90-x\right)+cos\left(180-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)

\(=cosx-cosx+sin^2x.\frac{1}{cos^2x}-tan^2x=tan^2x-tan^2x=0\)

Bài 1:

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

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

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

Bài 2:

Đề yêu cầu tính giá trị lượng giác nào bạn? sin?cos?tan?cot?

Không hỏi thì làm sao mà biết cần tính gì

tính giá trị lượng giác còn lại của góc \(\alpha\)

Giả sử các biểu thức đều xác định:

a/ \(sin^2x.tanx+cos^2x.cotx+2sinx.cosx\)

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

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

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

b/

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

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

\(=\frac{cosx}{cos^2x}=\frac{1}{cosx}\)

d/ \(\frac{sinx}{1+cosx}+\frac{1+cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}+\frac{sinx\left(1+cosx\right)}{sin^2x}\)

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

\(=\frac{2sinx}{sin^2x}=\frac{2}{sinx}\)

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

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

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

\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)

\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)

\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)

\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)

\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)

\(VT=tan^4x+cos^4x-2\left(tan^2x+cot^2x\right)+8\)

\(=\left(tan^2x+cot^2x\right)^2-2\left(tan^2x+cot^2x\right)+6\)

\(=\left(tan^2x+cot^2x-1\right)^2+5\)

Mặt khác áp dụng BĐT \(a^2+b^2\ge2ab\Rightarrow tan^2x+cot^2x\ge2\)

\(\Rightarrow\left(tan^2x+cot^2x-1\right)^2+5\ge\left(2-1\right)^2+5=6>5\Rightarrow VT>5\) (1)

Lại có \(3sinx-4cosx=5\left(sinx.\frac{3}{5}-cosx.\frac{4}{5}\right)\)

Do \(\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}\frac{3}{5}=cosa\\\frac{4}{5}=sina\end{matrix}\right.\)

\(\Rightarrow VP=3sinx-4cosx=5\left(sinx.cosa-cosx.sina\right)=5sin\left(x-a\right)\)

Do \(sin\left(x-a\right)\le1\Rightarrow5sin\left(x-a\right)\le5\Rightarrow VP\le5\) (2)

(1), (2) \(\Rightarrow VT>VP\)