a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)

\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)

\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)

\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )

\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)

Tham khảo:

a) 

Gọi M(x;y) là điểm trên đường tròn đơn vị sao cho \(\widehat {xOM} = \alpha \). Gọi N, P tương ứng là hình chiếu vuông góc của M lên các trục Ox, Oy.

Ta có: \(\left\{ \begin{array}{l}x = \cos \alpha \\y = \sin \alpha \end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}\alpha  = {x^2}\\{\sin ^2}\alpha  = {y^2}\end{array} \right.\)(1)

Mà \(\left\{ \begin{array}{l}\left| x \right| = ON\\\left| y \right| = OP = MN\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{x^2} = {\left| x \right|^2} = O{N^2}\\{y^2} = {\left| y \right|^2} = M{N^2}\end{array} \right.\)(2)

Từ (1) và (2) suy ra \({\sin ^2}\alpha  + {\cos ^2}\alpha  = O{N^2} + M{N^2} = O{M^2}\) (do \(\Delta OMN\) vuông tại N)

\( \Rightarrow {\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) (vì OM =1). (đpcm)

b) 

Ta có:  \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }}\;\;(\alpha  \ne {90^o})\)

\( \Rightarrow 1 + {\tan ^2}\alpha  = 1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}\)

Mà theo ý a) ta có \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) với mọi góc \(\alpha \)

\( \Rightarrow 1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\) (đpcm)

c) 

Ta có:  \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }}\;\;\;({0^o} < \alpha  < {180^o})\)

\( \Rightarrow 1 + {\cot ^2}\alpha  = 1 + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}\)

Mà theo ý a) ta có \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) với mọi góc \(\alpha \)

\( \Rightarrow 1 + {\cot ^2}\alpha  = \frac{1}{{{{\sin }^2}\alpha }}\) (đpcm)

\(=2\) luôn, sao còn để dòng cuối làm gì?

nhìn vào là không thấy gì cả

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

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

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

a) Ta có: \({\left( {\sin \alpha  + \cos \alpha } \right)^2} = {\sin ^2}\alpha  + 2\sin \alpha \cos \alpha  + {\cos ^2}\alpha  = 1 + \sin 2\alpha \;\)

b) \({\cos ^4}\alpha  - {\sin ^4}\alpha  = \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right) = \cos 2\alpha \;\)

\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)

\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)

\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)