=\(\frac{cosa}{sina}\)+\(\frac{sina}{cosa+1}\)

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

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

=\(\frac{1}{sina}\)

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

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

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

\(=2\)

b) Vẽ hình minh họa cho dễ nhìn nè :

A B C α

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

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

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

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

\(=sin^2\text{α}+cos^2\text{α}\)

\(=1\)

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

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

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

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

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

\(=sin^2\)

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

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

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

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

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

a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

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

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

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

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

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