a, \(\dfrac{1-sin2a}{1+sin2a}\)

\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)

\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)

\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)

b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)

\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)

\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)

\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)

\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)

\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)

.jkilfo,o7m5ijk

 Ta có $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alpha$

$=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)$

$=\sin \alpha .$

Vậy $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin \alpha$

1.

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

\(\Leftrightarrow\frac{1-2sin\alpha cos\alpha}{\left(sin\alpha-cos\alpha\right)\left(sin\alpha+cos\alpha\right)}=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)

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

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

\(\Leftrightarrow1-2sin\alpha cos\alpha=1-2sin\alpha cos\alpha\left(đpcm\right)\)

Bạn giúp mình bài này luôn với nha

Cho tam giác ABC ( AB < AC ) nội tiếp trong đường tròn (O) . Kẻ đường cao AH của tam giác ABC. Gọi P, Q lần lượt là chân đường vuông góc kẻ từ H xuống AB, AC .

1) Chứng minh rằng BCQP là tứ giác nội tiếp.

2) Hai đường thẳng BC,QP cắt nhau tại M . Chứng minh rằng: MH^2 = MB.MC .

3) Đường thẳng MA cắt đường tròn (O) tại K ( K khác A ). Gọi I là tâm đường tròn ngoại tiếp tứ giác BCQP . Chứng minh rằng I , H, K thẳng hàng.

1) \(\frac{1-2\sin\alpha\cdot\cos\alpha}{sin^2\alpha-\cos^2\alpha}=\frac{sin^2\alpha+\cos^2\alpha-2sin\alpha\cdot\cos\alpha}{sin^2\alpha-\cos^2\alpha}\)\(=\frac{\left(sin\alpha-\cos\alpha\right)^2}{sin^2\alpha-\cos^2\alpha}=\frac{sin\alpha-\cos\alpha}{sin\alpha+\cos\alpha}\)(đpcm)

2) \(cos^4\alpha+sin^2\alpha\cdot cos^2\alpha+sin^2\alpha\)

\(=cos^4\alpha+\left(1-cos^2\alpha\right)\cdot cos^2\alpha+sin^2\alpha\)

\(=cos^4\alpha+cos^2\alpha-cos^4\alpha+sin^2\alpha\)

\(=cos^2\alpha+sin^2\alpha=1\)(đpcm)

1)  \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)

bien doi ve phai ta co 

\(\frac{\sin\alpha}{\cos\alpha}=\frac{doi}{huyen}:\frac{ke}{huyen}=\frac{doi}{huyen}.\frac{huyen}{ke}=\frac{doi}{ke}=\tan\alpha\)

2) \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)

bien doi ve phai ta co

\(\frac{\cos\alpha}{\sin\alpha}=\frac{ke}{huyen}:\frac{doi}{huyen}=\frac{ke}{huyen}.\frac{huyen}{doi}=\frac{ke}{doi}=\cot\alpha\)

3) \(\tan\alpha.\cos\alpha=1\)

\(\frac{\cos\alpha}{\sin\alpha}.\frac{\sin\alpha}{\cos\alpha}=1\)

4) \(\sin^2\alpha+\cos^2\alpha=1\)

\(\frac{doi^2}{huyen^2}+\frac{ke^2}{huyen^2}=\frac{huyen^2}{huyen^2}=1\)( su dung dinh ly pitago ) 

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)

Ta có:

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

(đpcm)

b)

Ta có:

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

(đpcm)