a, = \(\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha\)+ \(\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\)

= \(2\sin^2\alpha+2\cos^2\alpha\)= 4

b,=\(\sin\alpha\cos\alpha\)(\(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\))

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

=1

#mã mã#

b, =1 

mk viết thiếu

#mã mã#

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

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

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

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

\(=1-sin^2a+sin^2a=1\)

a) \(\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

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

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

\(=2\)

b) Vẽ hình minh họa cho dễ nhìn nè :

\(sin\alpha\cdot cos\alpha\cdot\left(tan\alpha+cot\alpha\right)\)

\(=\frac{AC}{BC}\cdot\frac{AB}{BC}\cdot\left(\frac{AC}{AB}+\frac{AB}{AC}\right)\)

\(=\frac{AC\cdot AB\cdot AC}{BC\cdot BC\cdot AB}+\frac{AC\cdot AB\cdot AB}{BC\cdot BC\cdot AC}\)

\(=\left(\frac{AC}{BC}\right)^2+\left(\frac{AB}{BC}\right)^2\)

\(=sin^2\text{α}+cos^2\text{α}\)

\(=1\)

Lời giải:

Theo công thức lượng giác:

\(F=\sin (\pi +a)-\cos (\frac{\pi}{2}-a)+\cot (2\pi -a)+\tan (\frac{3\pi}{2}-a)\)

\(=-\sin a-\sin a+\cot (\pi -a)+\tan (\frac{\pi}{2}-a)\)

\(=-2\sin a-\cot a+\cot a=-2\sin a\)

Mình thay \(\alpha\) thành x để tiện ghi nhé

a) \(sinx.cosx\left(tanx+cotx\right)\)

\(=sinx.cosx\left(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\right)\)

\(=sinx.cosx\left(\dfrac{sinx^2+cosx^2}{sinx.cosx}\right)\)

\(=\dfrac{sinx.cosx}{sinx.cosx}=1\)

b) \(cot^2-cos^2.cot^2\)

\(=\dfrac{cos^2}{sin^2}-\left(1-sin^2\right).\dfrac{cos^2}{sin^2}\)

\(=\dfrac{cos^2-cos^2+sin^2cos^2}{sin^2}\)

\(=\dfrac{sin^2.cos^2}{sin^2}\)

\(=cos^2\)

c) \(tan^2-sin^2.tan^2\)

\(=tan^2\left(1-sin^2\right)\)

\(=\dfrac{sin^2}{cos^2}cos^2\)

\(=sin^2\)

a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)

\(=sin^2\alpha+cos^2\alpha=1\)

b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

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

c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)

\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)

e chưa nên k lm đk thông cảm cho e.

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

=\(cos^2a\)+\(sin^2a\)+\(sin^2a\)+\(cos^2a\)

=\(2sin^2a\)+\(2cos^2a\)

=\(2\left(sin^2a+cos^2a\right)\)

=2

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

=\(cos^2a+sin^2a+sin^2a+cos^2a\)

=\(2\left(sin^2a+cos^2a\right)\)=\(2.1=2\)