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

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

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

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

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

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

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

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

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

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

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

a) Ta có:  \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) =  - \cos {16^o}\end{array} \right.\)

\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)

b) 

Ta có:

\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) =  - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) =  - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) =  - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha  < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha  - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha  - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha  - \cos \alpha .\tan \alpha .\cot \alpha \)

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

`Q=sin^2 α +cot^2 α .sin^2 α`

`=sin^2 α + (cos^2 α)/(sin^2 α) .sin^2 α`

`=sin^2 α +cos^2 α`

`=1`.

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#

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

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

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

\(=sin^2\)