a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

- \(cosx=\frac{1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

- \(cosx=\frac{-1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

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

Bài 2:

\(0< a< \frac{\pi}{2}\Rightarrow sina>0\)

\(\Rightarrow sina=\sqrt{1-cos^2a}=\frac{\sqrt{3}}{2}\)

\(tana=\frac{sina}{cosa}=\sqrt{3}\)

\(cota=\frac{1}{tana}=\frac{1}{\sqrt{3}}\)

b/ \(-\frac{\pi}{2}< a< 0\Rightarrow cosa>0\)

\(\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{\sqrt{5}}{3}\)

\(tana=\frac{sina}{cosa}=-\frac{2\sqrt{5}}{5}\)

\(cota=\frac{1}{tana}=-\frac{\sqrt{5}}{2}\)

c/ \(-\frac{\pi}{2}< a< 0\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa>0\end{matrix}\right.\)

\(\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=...\)

Nhìn lại thì đề ảo quá, ai cho số \(\sqrt{\frac{3}{3}}\) bao giờ, như vầy người ta cho luôn số 1 cho rồi. Chắc bạn ghi nhầm, nhưng bạn có thể tự tính bằng cách thay vào công thức, sau đó tính \(sina=cosa.tana\)

d/ \(\frac{3\pi}{2}< 0< 2\pi\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa>0\end{matrix}\right.\)

\(\Rightarrow sina=-\frac{1}{\sqrt{1+cot^2a}}=-\frac{\sqrt{5}}{5}\)

\(cosa=sina.cota=\frac{2\sqrt{5}}{5}\)

\(tana=\frac{1}{cota}=-\frac{1}{2}\)

\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)

\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)

\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)

\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)

Giả sử tất cả các biểu thức đều xác định

a/

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

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

b/

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

c/

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

d/

\(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{1}{1+\frac{1}{tanx}}=\frac{1}{1+tanx}+\frac{tanx}{1+tanx}=\frac{1+tanx}{1+tanx}=1\)

e/

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

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

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)

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

\(=\left(1-sinx+cosx\right)^2\)

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)

\(A=s\left(x\right)cs\left(x\right)+\frac{\left(s^3\left(x\right)+cs^3\left(x\right)\right)}{cs\left(x\right)\left(1+t\left(x\right)\right)}=s\left(x\right)cs\left(x\right)+\left(\frac{\left(s\left(x\right)+cs\left(x\right)\right)\left(1-s\left(x\right)cs\left(x\right)\right)}{\left(s\left(x\right)+cs\left(x\right)\right)}\right)\)

\(=1\) vì \(s\left(x\right)+cs\left(x\right)\ne0,\forall0< =x< =\frac{\pi}{2}\)